2017年浙江工业大学大学生程序设计迎新赛预赛
2017-12-31 12:04
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这套题终于补完了。。。 嗨森阿
A
这是一个很简单的Nim游戏的博弈问题
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int num[qq];
int main(){
int t; scanf("%d", &t);
while (t--) {
int n, k; scanf("%d%d", &n, &k);
int ans = 0;
for (int i = 0; i < n; ++i) {
scanf("%d", num + i);
ans ^= num[i];
}
if (ans == 0 || (num[k - 1] < (num[k - 1] ^ ans))) {
puts("No");
} else {
puts("Yes");
}
}
return 0;
}
B
关键在于处理表达式，注意x + x这种情况
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1000000007;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
string st;
map<LL, LL> mp;
LL ct = 0;
void get(int &i) {
LL a, b, tmp = 1;
a = b = 0;
if (st[i] == '-') tmp = -1, ++i;
else if (st[i] == '+') ++i;
if (st[i] == 'x') {
a = tmp * 1;
} else if (isdigit(st[i])) {
while (isdigit(st[i])) {
a = a * 10 + st[i++] - '0';
}
a = a * tmp;
}
if (st[i] == 'x') {
i++;
if (st[i] == '^') {
i++;
while (isdigit(st[i])) {
b = b * 10 + st[i++] - '0';
}
} else {
b = 1;
}
}
if (mp.find(b) == mp.end()) {
mp[b] = a;
} else {
mp[b] += a;
}
}
LL quickPow(LL a, LL b) {
LL ans = 1;
while (b > 0) {
if (b & 1) ans = (ans * a) % MOD;
a = (a * a) % MOD;
b >>= 1;
}
return ans;
}
int main(){
ios::sync_with_stdio(false);
while (cin >> st) {
int len = st.size(), i = 0;
while (i < len) {
get(i);
}
map<LL, LL>::iterator it;
int t; cin >> t;
while (t--) {
LL x; cin >> x;
x %= MOD;
LL ans = ct;
for (it = mp.begin(); it != mp.end(); ++it) {
ans = (ans + (it->second * (quickPow(x, it->first)) % MOD + MOD) % MOD + MOD) % MOD;
}
cout << ans << endl;
}
mp.clear();
ct = 0;
}
return 0;
}
C
处理其中的映射关系，预处理前n项的阶乘，注意要求逆元
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
LL fin[qq];
LL quickPow(LL a, LL b) {
LL ans = 1;
while (b > 0) {
if(b & 1) ans = (ans * a) % MOD;
b >>= 1;
a = (a * a) % MOD;
}
return ans;
}
void Init() {
fin[0] = 1;
for (int i = 1; i < qq; ++i) {
fin[i] = (1LL * i * fin[i - 1]) % MOD;
}
}
map<LL, LL> mp;
LL num[qq];
LL n, m;
void solve(LL a, LL b) {
a = mp[a];
LL ans = (fin[b] * quickPow((fin[a] * fin[b - a]) % MOD, MOD - 2) ) % MOD;
printf("%lld\n", num[(ans % n) + 1]);
}
int main(){
Init();
int t; scanf("%d", &t);
while (t--) {
mp.clear();
scanf("%lld%lld", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%lld", num + i);
mp[num[i]] = i;
}
while (m--) {
LL a, b; scanf("%lld%lld", &a, &b);
solve(a, b);
}
}
return 0;
}
D
代码写的比较蠢，直接用的双端队列模拟的
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int tmp[qq];
void solve (int n, int m) {
deque<int> Q, tm;
deque<int>::iterator it;
int a, b;
int f = 0;
for (int i = 0; i < m; ++i) {
scanf("%d", &a);
if (a == 1) {
scanf("%d", &b);
if (!f) Q.push_front(b);
else Q.push_back(b);
} else if(a == 2) {
if (!f) Q.pop_front();
else Q.pop_back();
} else if(a == 3) {
scanf("%d", &b);
if (!f) Q.push_back(b);
else Q.push_front(b);
} else if(a == 4) {
if(!f) Q.pop_back();
else Q.pop_front();
} else if(a == 5) {
f ^= 1;
} else if(a == 6) {
printf("%d\n", (int)Q.size());
int cnt = 0;
for (it = Q.begin(); it != Q.end(); ++it) {
tmp[cnt++] = *it;
}
if (!f) {
for (int i = 0; i < cnt; ++i) {
printf("%d%c", tmp[i], i == cnt - 1 ? '\n' : ' ');
}
} else {
for (int i = cnt - 1; i >= 0; --i) {
printf("%d%c", tmp[i], i == 0 ? '\n' : ' ');
}
}
} else if(a == 7) {
int cnt = 0;
for (it = Q.begin(); it != Q.end(); ++it) {
tmp[cnt++] = *it;
}
Q.clear();
sort(tmp, tmp + cnt);
for(int i = 0; i < cnt; ++i) {
if(!f) Q.push_back(tmp[i]);
else Q.push_front(tmp[i]);
}
}
}
}
int main(){
int n, m; scanf("%d%d", &n, &m);
solve(n, m);
return 0;
}
E
二分答案，假设 总价值/总花费 >= x -> 总价值 - x * 总花费 >= 0， 所以我们可以按 总价值 -
x * 总花费来进行排序最后看结果是否大于0即可判断答案的可行性
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 6e6;
const LL INF = 1e9 + 10;
struct Node {
LL l, r;
}p[qq];
int n, k;
LL mid;
bool cmp(const Node &a, const Node &b) {
return a.l - mid * a.r > b.l - mid * b.r;
}
bool check() {
sort(p, p + n, cmp);
LL tmp = 0;
for (int i = 0; i < k; ++i) {
tmp += (p[i].l - mid * p[i].r);
}
// printf("%lld %lld\n", mid, tmp);
// printf("%lld\n", tmp >= 0);
if (tmp >= 0) return true;
return false;
}
int main(){
int t; scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &k);
for (int i = 0; i < n; ++i) {
scanf("%lld%lld", &p[i].r, &p[i].l);
}
LL l = 0, r = 1e8, ans = 0;
while (l <= r) {
mid = (l + r) >> 1;
if (check() == true) {
ans = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
printf("%lld\n", ans);
}
return 0;
}
F
思路还是简单的，跑一下最短路，然后依次判断小蝌蚪在第i秒的地方蝌蚪妈妈是否能到达，最后注意一下小蝌蚪不动的时候的蝌蚪妈妈到达的时间‘
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int n, m, k, s;
int p[qq], dis[qq];
bool vis[qq];
vector<int> G[qq];
void spfa() {
mst(vis, false);
queue<int> Q;
vis[s] = true;
mst(dis, 0x3f3f3f3f);
// printf("%d\n", dis[0]);
dis[s] = 0;
Q.push(s);
while (!Q.empty()) {
int u = Q.front();
Q.pop();
vis[u] = false;
for (int i = 0; i < (int)G[u].size(); ++i) {
int v = G[u][i];
if (dis[u] + 1 < dis[v]) {
dis[v] = dis[u] + 1;
if(vis[v]) continue;
vis[v] = true;
Q.push(v);
}
}
}
}
int main(){
while (scanf("%d%d%d%d", &n, &m, &k, &s) != EOF) {
for (int i = 1; i <= k; ++i) {
scanf("%d", p + i);
}
for (int a, b, i = 0; i < m; ++i) {
scanf("%d%d", &a, &b);
G[a].pb(b);
G[b].pb(a);
}
spfa();
int minx = INF;
for (int i = 1; i <= k; ++i) {
if(i >= dis[p[i]]) minx = min(minx, i);
}
minx = min(minx, max(dis[p[k]], k));
printf("%d\n", minx);
for (int i = 0; i <= n; ++i) {
G[i].clear();
}
}
return 0;
}
G
这题参考了题解的思路，维护一个单调递减的栈，每次输入一个高度，便和栈顶元素比较，如果栈顶元素小于当前高度，则这两个是可以互相监视的，在利用前缀和的思想，这一对是从栈顶元素所在的位置开始一直持续到当前元素的位置才结束，也就是说你的魔法屏障放在这两人中间的任何位置都可以消减这一对互相监视
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
const int MOD = 1e9 + 7;
int stk[qq][2];
int ans[qq];
int main() {
int t; scanf("%d", &t);
int Cas = 0;
while (t--) {
int top = 0;
int n; scanf("%d", &n);
mst(ans, 0);
for (int x, i = 1; i <= n; ++i) {
scanf("%d", &x);
if (!top) {
stk[++top][0] = i;
stk[top][1] = x;
} else {
while (top && stk[top][1] < x) {
ans[i]--;
ans[stk[top][0]]++;
top--;
}
if (top) {
ans[i]--;
ans[stk[top][0]]++;
}
stk[++top][0] = i;
stk[top][1] = x;
}
}
int maxn = 0, id;
for (int i = 1; i <= n; ++i) {
ans[i] += ans[i - 1];
if (ans[i] > maxn) {
maxn = ans[i];
id = i;
}
}
printf("Case #%d: %d %d\n", ++Cas, id + 1, maxn);
}
return 0;
}
H
这题需要对Havel-Hakim定理有所了解，关键还是在处理方面，看代码理解好了
Havel-Hakim定理
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int maxn[qq * 4], lazy[qq * 4], a[qq];
void pushUp(int rt) {
maxn[rt] = max(maxn[lson], maxn[rson]);
}
void pushDown(int rt) {
if (lazy[rt]) {
maxn[lson] += lazy[rt];
maxn[rson] += lazy[rt];
lazy[lson] += lazy[rt];
lazy[rson] += lazy[rt];
lazy[rt] = 0;
}
}
void build(int l, int r, int rt) {
lazy[rt] = 0;
if (l == r) {
maxn[rt] = a[l];
return;
}
int mid = (l + r) >> 1;
build(l, mid, lson);
build(mid + 1, r, rson);
pushUp(rt);
}
int first(int l, int r, int rt) {
if (l == r) {
return l;
}
pushDown(rt);
int mid = (l + r) >> 1;
int res;
if (maxn[lson] > 0) res = first(l, mid, lson);
else res = first(mid + 1, r, rson);
pushUp(rt);
return res;
}
int get(int p, int l, int r, int rt) {
if (l == r) {
return maxn[rt];
}
pushDown(rt);
int res;
int mid = (l + r) >> 1;
if (p <= mid) res = get(p, l, mid, lson);
else res = get(p, mid + 1, r, rson);
pushUp(rt);
return res;
}
void upDate(int x, int y, int val, int l, int r, int rt) {
if (x <= l && r <= y) {
maxn[rt] += val;
lazy[rt] += val;
return;
}
pushDown(rt);
int mid = (l + r) >> 1;
if (x <= mid) upDate(x, y, val, l, mid, lson);
if (y > mid) upDate(x, y, val, mid + 1, r, rson);
pushUp(rt);
}
int main(){
int t; scanf("%d", &t);
while (t--) {
int n; scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", a + i);
}
sort(a + 1, a + 1 + n);
build(1, n, 1);
for (int i = n; i >= 1; --i) {
int y = first(1, n, 1);
if (y >= i) break;
int x = get(i, 1, n, 1);
if (x >= i - y) {
upDate(y, i - 1, -1, 1, n, 1);
upDate(i, i, -(i - y), 1, n, 1);
} else {
upDate(i, i, -x, 1, n, 1);
upDate(i - x, i - 1, -1, 1, n, 1);
if (get(i - x - 1, 1, n, 1) > get(i - x, 1, n, 1)) {
int L1 = i - x - 1, R1 = i - x - 1, L2 = i - x, R2 = i - x;
int left = y, right = i - x - 1;
while (left <= right) {
int mid = (left + right) >> 1;
if (get(mid, 1, n, 1) < get(i - x - 1, 1, n, 1)) {
left = mid + 1;
} else {
right = mid - 1;
L1 = mid;
}
}
left = i - x, right = i - 1;
while (left <= right) {
int mid = (left + right) >> 1;
if (get(mid, 1, n, 1) > get(i - x, 1, n, 1)) {
right = mid - 1;
} else {
left = mid + 1;
R2 = mid;
}
}
if (R1 - L1 == R2 - L2) {
upDate(L1, R1, -1, 1, n, 1);
upDate(L2, R2, 1, 1, n, 1);
} else if (R1 - L1 > R2 - L2) {
upDate(L2, R2, 1, 1, n, 1);
upDate(L1, L1 + (R2 - L2), -1, 1, n, 1);
} else {
upDate(L1, R1, -1, 1, n, 1);
upDate(R2 - (R1 - L1), R2, 1, 1, n, 1);
}
}
}
/*for (int j = 1; j <= i; ++j) {
printf("%d ", get(j, 1, n, 1));
}
puts("");*/
}
LL ans = 0;
for (int i = 1; i <= n; ++i) {
// printf("%d\n", get(i, 1, n, 1));
ans += (LL)get(i, 1, n, 1);
}
printf("%lld\n", ans);
}
return 0;
}
I
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int num[qq];
int main(){
int n; scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
if(i % 4 == 0) continue;
bool f = false;
int tm = i;
while (tm > 0) {
if(tm % 10 == 4) f = true;
tm /= 10;
}
if(f) continue;
printf("%d\n", i);
}
return 0;
J
这题在图书画一画就明白了
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e3 + 10;
const int INF = 1e9 + 10;
const double PI = 4.0 * atan(1.0);
double r, c, R;
int n;
double x[qq], y[qq];
double check(double a) {
int t = a;
if (t == 0) {
double tm = a;
for (int i = 0; i < 2; ++i) {
tm *= 10;
int b = tm;
b %= 10;
if(b != 0) return a;
}
return 0;
}
return a;
}
int main(){
int t; scanf("%d", &t);
while (t--) {
scanf("%lf%lf%lf%d", &r, &c, &R, &n);
double gap = 2.0 * PI / n;
double init = 0.0;
for (int i = 0; i < n
e370
; ++i) {
x[i] = R * cos(init);
y[i] = R * sin(init);
init += gap;
}
printf("%.2lf %.2lf\n", check(x[0] + r), check(y[0] + c));
for (int i = n - 1; i >= 1; --i) {
printf("%.2lf %.2lf\n", check(x[i] + r), check(y[i] + c));
}
}
return 0;
}
K
这题我居然直接跑了最小生成树。。。。。
其实只需要将所有边连接到点权制最大的那个点即可
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e3 + 10;
const int INF = 1e9 + 10;
int mr[qq][qq];
int num[qq], dis[qq], n;
bool vis[qq];
void solve() {
int ans = 0;
mst(vis, false);
vis[1] = true;
for (int i = 1; i <= n; ++i) {
dis[i] = mr[1][i];
}
for (int i = 1; i < n; ++i) {
int maxn = -1, k;
for (int j = 1; j <= n; ++j) {
if(vis[j]) continue;
if(maxn < dis[j]) maxn = dis[k = j];
}
vis[k] = true;
ans += maxn;
for (int j = 1; j <= n; ++j) {
if(vis[j]) continue;
dis[j] = max(dis[j], mr[k][j]);
}
}
printf("%d\n", ans);
}
int main(){
int t; scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", num + i);
for(int j = 1; j < i; ++j) {
mr[i][j] = mr[j][i] = (num[i] + num[j]) / 2;
}
}
solve();
}
return 0;
}
L
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int num[qq];
int main(){
int t; scanf("%d", &t);
while (t--) {
for (int i = 0; i < 3; ++i) {
scanf("%d", num + i);
}
int a = max(num[0], max(num[1], num[2]));
int c = min(num[0], min(num[1], num[2]));
int b = num[0] + num[1] + num[2] - a - c;
printf("%d\n", a * 100 + b * 10 + c);
}
return 0;
}
M
这题我用队列模拟了一下，有些特殊情况还是需要考虑的
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int num[qq];
int main(){
int t; scanf("%d", &t);
while (t--) {
int n; scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", num + i);
}
int b; scanf("%d", &b);
sort(num, num + n);
priority_queue<int, vector<int>, greater<int> > Q;
for (int i = 0; i < n; ++i) {
if (Q.size() < 3) {
Q.push(num[i] + b);
} else {
int u = Q.top();
Q.pop();
if (u <= num[i]) {
Q.push(num[i] + b);
} else {
Q.push(u + b);
}
}
}
int maxn = 0;
while (!Q.empty()) {
maxn = max(maxn, Q.top());
Q.pop();
}
printf("%d\n", maxn);
}
return 0;
}
A
这是一个很简单的Nim游戏的博弈问题
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int num[qq];
int main(){
int t; scanf("%d", &t);
while (t--) {
int n, k; scanf("%d%d", &n, &k);
int ans = 0;
for (int i = 0; i < n; ++i) {
scanf("%d", num + i);
ans ^= num[i];
}
if (ans == 0 || (num[k - 1] < (num[k - 1] ^ ans))) {
puts("No");
} else {
puts("Yes");
}
}
return 0;
}
B
关键在于处理表达式，注意x + x这种情况
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1000000007;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
string st;
map<LL, LL> mp;
LL ct = 0;
void get(int &i) {
LL a, b, tmp = 1;
a = b = 0;
if (st[i] == '-') tmp = -1, ++i;
else if (st[i] == '+') ++i;
if (st[i] == 'x') {
a = tmp * 1;
} else if (isdigit(st[i])) {
while (isdigit(st[i])) {
a = a * 10 + st[i++] - '0';
}
a = a * tmp;
}
if (st[i] == 'x') {
i++;
if (st[i] == '^') {
i++;
while (isdigit(st[i])) {
b = b * 10 + st[i++] - '0';
}
} else {
b = 1;
}
}
if (mp.find(b) == mp.end()) {
mp[b] = a;
} else {
mp[b] += a;
}
}
LL quickPow(LL a, LL b) {
LL ans = 1;
while (b > 0) {
if (b & 1) ans = (ans * a) % MOD;
a = (a * a) % MOD;
b >>= 1;
}
return ans;
}
int main(){
ios::sync_with_stdio(false);
while (cin >> st) {
int len = st.size(), i = 0;
while (i < len) {
get(i);
}
map<LL, LL>::iterator it;
int t; cin >> t;
while (t--) {
LL x; cin >> x;
x %= MOD;
LL ans = ct;
for (it = mp.begin(); it != mp.end(); ++it) {
ans = (ans + (it->second * (quickPow(x, it->first)) % MOD + MOD) % MOD + MOD) % MOD;
}
cout << ans << endl;
}
mp.clear();
ct = 0;
}
return 0;
}
C
处理其中的映射关系，预处理前n项的阶乘，注意要求逆元
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
LL fin[qq];
LL quickPow(LL a, LL b) {
LL ans = 1;
while (b > 0) {
if(b & 1) ans = (ans * a) % MOD;
b >>= 1;
a = (a * a) % MOD;
}
return ans;
}
void Init() {
fin[0] = 1;
for (int i = 1; i < qq; ++i) {
fin[i] = (1LL * i * fin[i - 1]) % MOD;
}
}
map<LL, LL> mp;
LL num[qq];
LL n, m;
void solve(LL a, LL b) {
a = mp[a];
LL ans = (fin[b] * quickPow((fin[a] * fin[b - a]) % MOD, MOD - 2) ) % MOD;
printf("%lld\n", num[(ans % n) + 1]);
}
int main(){
Init();
int t; scanf("%d", &t);
while (t--) {
mp.clear();
scanf("%lld%lld", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%lld", num + i);
mp[num[i]] = i;
}
while (m--) {
LL a, b; scanf("%lld%lld", &a, &b);
solve(a, b);
}
}
return 0;
}
D
代码写的比较蠢，直接用的双端队列模拟的
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int tmp[qq];
void solve (int n, int m) {
deque<int> Q, tm;
deque<int>::iterator it;
int a, b;
int f = 0;
for (int i = 0; i < m; ++i) {
scanf("%d", &a);
if (a == 1) {
scanf("%d", &b);
if (!f) Q.push_front(b);
else Q.push_back(b);
} else if(a == 2) {
if (!f) Q.pop_front();
else Q.pop_back();
} else if(a == 3) {
scanf("%d", &b);
if (!f) Q.push_back(b);
else Q.push_front(b);
} else if(a == 4) {
if(!f) Q.pop_back();
else Q.pop_front();
} else if(a == 5) {
f ^= 1;
} else if(a == 6) {
printf("%d\n", (int)Q.size());
int cnt = 0;
for (it = Q.begin(); it != Q.end(); ++it) {
tmp[cnt++] = *it;
}
if (!f) {
for (int i = 0; i < cnt; ++i) {
printf("%d%c", tmp[i], i == cnt - 1 ? '\n' : ' ');
}
} else {
for (int i = cnt - 1; i >= 0; --i) {
printf("%d%c", tmp[i], i == 0 ? '\n' : ' ');
}
}
} else if(a == 7) {
int cnt = 0;
for (it = Q.begin(); it != Q.end(); ++it) {
tmp[cnt++] = *it;
}
Q.clear();
sort(tmp, tmp + cnt);
for(int i = 0; i < cnt; ++i) {
if(!f) Q.push_back(tmp[i]);
else Q.push_front(tmp[i]);
}
}
}
}
int main(){
int n, m; scanf("%d%d", &n, &m);
solve(n, m);
return 0;
}
E
二分答案，假设 总价值/总花费 >= x -> 总价值 - x * 总花费 >= 0， 所以我们可以按 总价值 -
x * 总花费来进行排序最后看结果是否大于0即可判断答案的可行性
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 6e6;
const LL INF = 1e9 + 10;
struct Node {
LL l, r;
}p[qq];
int n, k;
LL mid;
bool cmp(const Node &a, const Node &b) {
return a.l - mid * a.r > b.l - mid * b.r;
}
bool check() {
sort(p, p + n, cmp);
LL tmp = 0;
for (int i = 0; i < k; ++i) {
tmp += (p[i].l - mid * p[i].r);
}
// printf("%lld %lld\n", mid, tmp);
// printf("%lld\n", tmp >= 0);
if (tmp >= 0) return true;
return false;
}
int main(){
int t; scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &k);
for (int i = 0; i < n; ++i) {
scanf("%lld%lld", &p[i].r, &p[i].l);
}
LL l = 0, r = 1e8, ans = 0;
while (l <= r) {
mid = (l + r) >> 1;
if (check() == true) {
ans = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
printf("%lld\n", ans);
}
return 0;
}
F
思路还是简单的，跑一下最短路，然后依次判断小蝌蚪在第i秒的地方蝌蚪妈妈是否能到达，最后注意一下小蝌蚪不动的时候的蝌蚪妈妈到达的时间‘
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int n, m, k, s;
int p[qq], dis[qq];
bool vis[qq];
vector<int> G[qq];
void spfa() {
mst(vis, false);
queue<int> Q;
vis[s] = true;
mst(dis, 0x3f3f3f3f);
// printf("%d\n", dis[0]);
dis[s] = 0;
Q.push(s);
while (!Q.empty()) {
int u = Q.front();
Q.pop();
vis[u] = false;
for (int i = 0; i < (int)G[u].size(); ++i) {
int v = G[u][i];
if (dis[u] + 1 < dis[v]) {
dis[v] = dis[u] + 1;
if(vis[v]) continue;
vis[v] = true;
Q.push(v);
}
}
}
}
int main(){
while (scanf("%d%d%d%d", &n, &m, &k, &s) != EOF) {
for (int i = 1; i <= k; ++i) {
scanf("%d", p + i);
}
for (int a, b, i = 0; i < m; ++i) {
scanf("%d%d", &a, &b);
G[a].pb(b);
G[b].pb(a);
}
spfa();
int minx = INF;
for (int i = 1; i <= k; ++i) {
if(i >= dis[p[i]]) minx = min(minx, i);
}
minx = min(minx, max(dis[p[k]], k));
printf("%d\n", minx);
for (int i = 0; i <= n; ++i) {
G[i].clear();
}
}
return 0;
}
G
这题参考了题解的思路，维护一个单调递减的栈，每次输入一个高度，便和栈顶元素比较，如果栈顶元素小于当前高度，则这两个是可以互相监视的，在利用前缀和的思想，这一对是从栈顶元素所在的位置开始一直持续到当前元素的位置才结束，也就是说你的魔法屏障放在这两人中间的任何位置都可以消减这一对互相监视
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
const int MOD = 1e9 + 7;
int stk[qq][2];
int ans[qq];
int main() {
int t; scanf("%d", &t);
int Cas = 0;
while (t--) {
int top = 0;
int n; scanf("%d", &n);
mst(ans, 0);
for (int x, i = 1; i <= n; ++i) {
scanf("%d", &x);
if (!top) {
stk[++top][0] = i;
stk[top][1] = x;
} else {
while (top && stk[top][1] < x) {
ans[i]--;
ans[stk[top][0]]++;
top--;
}
if (top) {
ans[i]--;
ans[stk[top][0]]++;
}
stk[++top][0] = i;
stk[top][1] = x;
}
}
int maxn = 0, id;
for (int i = 1; i <= n; ++i) {
ans[i] += ans[i - 1];
if (ans[i] > maxn) {
maxn = ans[i];
id = i;
}
}
printf("Case #%d: %d %d\n", ++Cas, id + 1, maxn);
}
return 0;
}
H
这题需要对Havel-Hakim定理有所了解，关键还是在处理方面，看代码理解好了
Havel-Hakim定理
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int maxn[qq * 4], lazy[qq * 4], a[qq];
void pushUp(int rt) {
maxn[rt] = max(maxn[lson], maxn[rson]);
}
void pushDown(int rt) {
if (lazy[rt]) {
maxn[lson] += lazy[rt];
maxn[rson] += lazy[rt];
lazy[lson] += lazy[rt];
lazy[rson] += lazy[rt];
lazy[rt] = 0;
}
}
void build(int l, int r, int rt) {
lazy[rt] = 0;
if (l == r) {
maxn[rt] = a[l];
return;
}
int mid = (l + r) >> 1;
build(l, mid, lson);
build(mid + 1, r, rson);
pushUp(rt);
}
int first(int l, int r, int rt) {
if (l == r) {
return l;
}
pushDown(rt);
int mid = (l + r) >> 1;
int res;
if (maxn[lson] > 0) res = first(l, mid, lson);
else res = first(mid + 1, r, rson);
pushUp(rt);
return res;
}
int get(int p, int l, int r, int rt) {
if (l == r) {
return maxn[rt];
}
pushDown(rt);
int res;
int mid = (l + r) >> 1;
if (p <= mid) res = get(p, l, mid, lson);
else res = get(p, mid + 1, r, rson);
pushUp(rt);
return res;
}
void upDate(int x, int y, int val, int l, int r, int rt) {
if (x <= l && r <= y) {
maxn[rt] += val;
lazy[rt] += val;
return;
}
pushDown(rt);
int mid = (l + r) >> 1;
if (x <= mid) upDate(x, y, val, l, mid, lson);
if (y > mid) upDate(x, y, val, mid + 1, r, rson);
pushUp(rt);
}
int main(){
int t; scanf("%d", &t);
while (t--) {
int n; scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", a + i);
}
sort(a + 1, a + 1 + n);
build(1, n, 1);
for (int i = n; i >= 1; --i) {
int y = first(1, n, 1);
if (y >= i) break;
int x = get(i, 1, n, 1);
if (x >= i - y) {
upDate(y, i - 1, -1, 1, n, 1);
upDate(i, i, -(i - y), 1, n, 1);
} else {
upDate(i, i, -x, 1, n, 1);
upDate(i - x, i - 1, -1, 1, n, 1);
if (get(i - x - 1, 1, n, 1) > get(i - x, 1, n, 1)) {
int L1 = i - x - 1, R1 = i - x - 1, L2 = i - x, R2 = i - x;
int left = y, right = i - x - 1;
while (left <= right) {
int mid = (left + right) >> 1;
if (get(mid, 1, n, 1) < get(i - x - 1, 1, n, 1)) {
left = mid + 1;
} else {
right = mid - 1;
L1 = mid;
}
}
left = i - x, right = i - 1;
while (left <= right) {
int mid = (left + right) >> 1;
if (get(mid, 1, n, 1) > get(i - x, 1, n, 1)) {
right = mid - 1;
} else {
left = mid + 1;
R2 = mid;
}
}
if (R1 - L1 == R2 - L2) {
upDate(L1, R1, -1, 1, n, 1);
upDate(L2, R2, 1, 1, n, 1);
} else if (R1 - L1 > R2 - L2) {
upDate(L2, R2, 1, 1, n, 1);
upDate(L1, L1 + (R2 - L2), -1, 1, n, 1);
} else {
upDate(L1, R1, -1, 1, n, 1);
upDate(R2 - (R1 - L1), R2, 1, 1, n, 1);
}
}
}
/*for (int j = 1; j <= i; ++j) {
printf("%d ", get(j, 1, n, 1));
}
puts("");*/
}
LL ans = 0;
for (int i = 1; i <= n; ++i) {
// printf("%d\n", get(i, 1, n, 1));
ans += (LL)get(i, 1, n, 1);
}
printf("%lld\n", ans);
}
return 0;
}
I
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int num[qq];
int main(){
int n; scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
if(i % 4 == 0) continue;
bool f = false;
int tm = i;
while (tm > 0) {
if(tm % 10 == 4) f = true;
tm /= 10;
}
if(f) continue;
printf("%d\n", i);
}
return 0;
J
这题在图书画一画就明白了
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e3 + 10;
const int INF = 1e9 + 10;
const double PI = 4.0 * atan(1.0);
double r, c, R;
int n;
double x[qq], y[qq];
double check(double a) {
int t = a;
if (t == 0) {
double tm = a;
for (int i = 0; i < 2; ++i) {
tm *= 10;
int b = tm;
b %= 10;
if(b != 0) return a;
}
return 0;
}
return a;
}
int main(){
int t; scanf("%d", &t);
while (t--) {
scanf("%lf%lf%lf%d", &r, &c, &R, &n);
double gap = 2.0 * PI / n;
double init = 0.0;
for (int i = 0; i < n
e370
; ++i) {
x[i] = R * cos(init);
y[i] = R * sin(init);
init += gap;
}
printf("%.2lf %.2lf\n", check(x[0] + r), check(y[0] + c));
for (int i = n - 1; i >= 1; --i) {
printf("%.2lf %.2lf\n", check(x[i] + r), check(y[i] + c));
}
}
return 0;
}
K
这题我居然直接跑了最小生成树。。。。。
其实只需要将所有边连接到点权制最大的那个点即可
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e3 + 10;
const int INF = 1e9 + 10;
int mr[qq][qq];
int num[qq], dis[qq], n;
bool vis[qq];
void solve() {
int ans = 0;
mst(vis, false);
vis[1] = true;
for (int i = 1; i <= n; ++i) {
dis[i] = mr[1][i];
}
for (int i = 1; i < n; ++i) {
int maxn = -1, k;
for (int j = 1; j <= n; ++j) {
if(vis[j]) continue;
if(maxn < dis[j]) maxn = dis[k = j];
}
vis[k] = true;
ans += maxn;
for (int j = 1; j <= n; ++j) {
if(vis[j]) continue;
dis[j] = max(dis[j], mr[k][j]);
}
}
printf("%d\n", ans);
}
int main(){
int t; scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", num + i);
for(int j = 1; j < i; ++j) {
mr[i][j] = mr[j][i] = (num[i] + num[j]) / 2;
}
}
solve();
}
return 0;
}
L
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int num[qq];
int main(){
int t; scanf("%d", &t);
while (t--) {
for (int i = 0; i < 3; ++i) {
scanf("%d", num + i);
}
int a = max(num[0], max(num[1], num[2]));
int c = min(num[0], min(num[1], num[2]));
int b = num[0] + num[1] + num[2] - a - c;
printf("%d\n", a * 100 + b * 10 + c);
}
return 0;
}
M
这题我用队列模拟了一下，有些特殊情况还是需要考虑的
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
#define pill pair<int, int>
#define mst(a, b) memset(a, b, sizeof a)
#define REP(i, x, n) for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
int num[qq];
int main(){
int t; scanf("%d", &t);
while (t--) {
int n; scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", num + i);
}
int b; scanf("%d", &b);
sort(num, num + n);
priority_queue<int, vector<int>, greater<int> > Q;
for (int i = 0; i < n; ++i) {
if (Q.size() < 3) {
Q.push(num[i] + b);
} else {
int u = Q.top();
Q.pop();
if (u <= num[i]) {
Q.push(num[i] + b);
} else {
Q.push(u + b);
}
}
}
int maxn = 0;
while (!Q.empty()) {
maxn = max(maxn, Q.top());
Q.pop();
}
printf("%d\n", maxn);
}
return 0;
}
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