2017 Multi-University Training Contest - Team 3

1003

题意：给出n和k，然后给出n个数，求所有区间第k大数的和，区间长度小于k则是0

思路：参考了题解的思路，枚举第k大数是x，维护一个链表，链表里面全部是>=x的数，同时向左向右走k次，然后计算答案

虽然代码不长，但感觉其中很多东西都得想清楚

求答案过程实际上是要求左边0个比x大的，右边k - 1个比x大，然后左边1个比x大的，右边k - 2个比x大的，依次类推

#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>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#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 = 5e5 + 10;
int pos[qq], pre[qq], suf[qq];
int x[105], y[105];
int a, b;
int n, k;
void Erase(int x) {
int pp = pre[x];
int nn = suf[x];
if(pp)	suf[pre[x]] = nn;
if(nn <= n)	pre[suf[x]] = pp;
pre[x] = suf[x] = 0;
}

int main(){
int t;	scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &k);
for(int num, i = 1; i <= n; ++i) {
scanf("%d", &num);
pos[num] = i;
}
for(int i = 1; i <= n; ++i) {
pre[i] = suf[i] = 0;
}
for(int i = 1; i <= n; ++i) {
pre[i] = i - 1, suf[i] = i + 1;
}
LL ans = 0;
for(int num = 1; num <= n - k + 1; ++num) {
a = b = 0;
int p = pos[num];
for(int d = p; d != 0 && a <= k + 1; d = pre[d])	x[++a] = d;
for(int d = p; d != n + 1 && b <= k + 1; d = suf[d])	y[++b] = d;
//			printf("%d %d\n", a, b);
x[++a] = 0;
y[++b] = n + 1;
for(int i = 1; i <= a - 1; ++i) {
if(k + 1 - i >= 1 && k + 1 - i + 1 <= b){
ans += (LL)1LL * (x[i] - x[i + 1]) * (y[k + 1 - i + 1] - y[k + 1 - i]) * num;
}
}
Erase(p);
}
printf("%lld\n", ans);
}
return 0;
}

1005

题意：给出n个点n - 1条边组成的树，每条边有权值，现在问你把2 3 ... n 这n - 1个点分成k部分，这k部分都是两两不相交的，然后在k部分中每一部分添加1这个点，然后求出每部分的最小生成树，要求k部分的最小生成树的权值和最大，求出这个值

思路：以1为根建树，算每一条边的贡献，如果这条边下面有x个结点，如果x >= k那么这条边会被算k次，否则就是x次，相当于把这些点放到不同的集合中去，这条边就可以多算几次

#include <cstdio>
#include <cstring>
#include <cmath>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define ft first
#define sd second
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int qq = 1e6 + 10;
const int MOD = 1e9 + 7;
LL num[qq];
vector<int> G[qq], dis[qq];
LL n, k;
LL ans = 0;
void Dfs(int u, LL dep, int fa) {
int len = G[u].size();
for(int i = 0; i < len; ++i) {
int v = G[u][i];
if(v == fa)	continue;
Dfs(v, dep + dis[u][i], u);
if(num[v] >= k)	ans += (LL)dis[u][i] * k;
else	ans += (LL)dis[u][i] * num[v];
num[u] += num[v];
}
num[u] += 1;
}
void Init() {
mst(num, 0);
for(int i = 0; i <= n; ++i) {
G[i].clear(), dis[i].clear();
}
ans = 0;
}

int main(){
while(scanf("%lld%lld", &n, &k) != EOF) {
Init();
for(int i = 1; i < n; ++i) {
int x, y, z;	scanf("%d%d%d", &x, &y, &z);
G[x].pb(y), dis[x].pb(z);
G[y].pb(x), dis[y].pb(z);
}
Dfs(1, 0, -1);
printf("%lld\n", ans);
}
return 0;
}

1008

思路：手算了几个猜了一下、结果就是求n^k

#include <cstdio>
#include <cstring>
#include <cmath>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define ft first
#define sd second
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int qq = 1e5 + 10;
const int MOD = 1e9 + 7;
LL Solve(LL n, LL k) {
LL ans = 1;
while(k > 0) {
if(k & 1)	ans = (ans * n) % MOD;
n = (n * n) % MOD;
k >>= 1;
}
return ans;
}

int main(){
LL n, k;
int Cas = 0;
while(scanf("%lld%lld", &n, &k) != EOF) {
n %= MOD;
printf("Case #%d: %lld\n", ++Cas, Solve(n, k));
}
return 0;
}

1011

签到

#include <cstdio>
#include <cstring>
#include <cmath>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define ft first
#define sd second
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int qq = 1e5 + 10;
const int MOD = 998244353;

int main(){
int t;	scanf("%d", &t);
int cnt = 0;
while(t--) {
int x;	scanf("%d", &x);
if(x <= 35)	cnt++;
}
printf("%d\n", cnt);
return 0;
}