Codeforces Round #264 (Div. 2)

Codeforces Round #264 (Div. 2)

题目链接

A：注意特判正好的情况，其他就一个个去判断记录最大值即可

B：扫一遍，不够的用钱去填即可，把多余能量记录下来

C：把主副对角线处理出来，然后黑格白格只能各选一个最大的放即可

D：转化为DAG最长路问题，每个数字记录下在每个序列的位置，如果一个数字能放上去，那么肯定是每个序列上的数字都是在之前最末尾数字的后面

E：大力出奇迹，预处理出树，然后每次查询从当前位置一直往上找到一个符合的即可，如果没有符合的就是-1

代码：

A：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int n, s;

int main() {
scanf("%d%d", &n, &s);
int x, y;
int flag = 1;
int ans = 0;
for (int i = 0; i < n; i++) {
scanf("%d%d", &x, &y);
if (x == s) {
if (y == 0) {
ans = max(ans, y);
flag = 0;
}
}
else if (x < s) {
ans = max(ans, (100 - y) % 100);
flag = 0;
}
}
if (flag) printf("-1\n");
else printf("%d\n", ans);
return 0;
}

B：

#include <cstdio>
#include <cstring>

const int N = 100005;
typedef long long ll;

int n;
ll h
;

int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%lld", &h[i]);
ll now = 0;
ll ans = 0;
for (int i = 1; i <= n; i++) {
if (h[i] > h[i - 1]) {
ll need = h[i] - h[i - 1];
if (now >= need) {
now -= need;
} else {
ans += need - now;
now = 0;
}
} else {
now += h[i - 1] - h[i];
}
}
printf("%lld\n", ans);
return 0;
}

C：

#include <cstdio>
#include <cstring>

const int N = 2005;

typedef long long ll;

int n;
ll g

, zhu[N + N], fu[N + N];

int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
scanf("%lld", &g[i][j]);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
zhu[i - j + n] += g[i][j];
fu[i + j] += g[i][j];
}
}
ll b = -1, w = -1;
int bx, by, wx, wy;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
ll sum = zhu[i - j + n] + fu[i + j] - g[i][j];
if ((i + j) % 2 == 0) {
if (b < sum) {
b = sum;
bx = i + 1;
by = j + 1;
}
}
else {
if (w < sum) {
w = sum;
wx = i + 1;
wy = j + 1;
}
}
}
}
printf("%lld\n", b + w);
printf("%d %d %d %d\n", bx, by, wx, wy);
return 0;
}

D：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1005;

struct Num {
int v[10], la;
} num
;

bool cmp(Num a, Num b) {
return a.la < b.la;
}

int n, k, dp

;

bool judge(int i, int j) {
for (int x = 0; x < k; x++) {
if (num[i].v[x] < num[j].v[x]) return false;
}
return true;
}

int main() {
scanf("%d%d", &n, &k);
for (int i = 0; i < k; i++) {
int tmp;
for (int j = 1; j <= n; j++) {
scanf("%d", &tmp);
num[tmp].v[i] = j;
}
}
for (int i = 0; i <= n; i++) {
int maxv = 0;
for (int j = 0; j < k; j++) {
maxv = max(maxv, num[i].v[j]);
}
num[i].la = maxv;
}
sort(num, num + n + 1, cmp);
/*
for (int i = 0; i <= n; i++) {
for (int j = 0; j < k; j++) {
printf("%d ", num[i].v[j]);
}
printf("\n");
}*/
for (int i = 1; i <= n; i++) {
for (int j = 0; j < i; j++) {
dp[i][j] = max(dp[i][j], dp[i - 1][j]);
if (judge(i, j)) {
dp[i][i] = max(dp[i][i], dp[i][j] + 1);
}
}
}
int ans = 0;
for (int i = 0; i <= n; i++)
ans = max(ans, dp
[i]);
printf("%d\n", ans);
return 0;
}

E：

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

const int N = 100005;

int n, q, val
, f
;
vector<int> g
;

void dfs(int u, int fa) {
f[u] = fa;
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (v == fa) continue;
dfs(v, u);
}
}

int gcd(int a, int b) {
while (b) {
int tmp = a % b;
a = b;
b = tmp;
}
return a;
}

int query(int u) {
int v = val[u];
u = f[u];
while (u) {
if (gcd(val[u], v) > 1) return u;
u = f[u];
}
return -1;
}

int main() {
scanf("%d%d", &n, &q);
for (int i = 1; i <= n; i++)
scanf("%d", &val[i]);
int u, v;
for (int i = 1; i < n; i++) {
scanf("%d%d", &u, &v);
g[u].push_back(v);
g[v].push_back(u);
}
dfs(1, 0);
int tp, a, b;
while (q--) {
scanf("%d%d", &tp, &a);
if (tp == 1) {
printf("%d\n", query(a));
} else {
scanf("%d", &b);
val[a] = b;
}
}
return 0;
}