Manthan, Codefest 17

A

#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)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e6 + 10;const int INF = 1e9 + 10;map<string, bool> mp;int main(){int n;cin >> n;string st;for(int i = 0; i < n; ++i) {cin >> st;if(mp[st])puts("YES");elseputs("NO");mp[st] = true;}return 0;}

B

题意：给n个数，再给你p，q，r，要求你选择i、j、k(i <= j <= k)，a[i] * p + a[j] * q + a[k] * r最大

思路：很显然存在前后关系，我们可以对a[i] * p做一个前缀，对a[k] * r 做一个后缀，然后枚举a[j] * q

#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)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e6 + 10;const LL INF = 3e18 + 10;LL p, q, r, n;LL num[qq], pre[qq], suf[qq];int main(){cin >> n >> p >> q >> r;for(int i = 1; i <= n; ++i) {cin >> num[i];}for(int i = 1; i <= n; ++i) {if(i == 1)pre[i] = num[i] * p;elsepre[i] = max(pre[i - 1], p * num[i]);}for(int i = n; i >= 1; --i) {if(i == n)suf[i] = num[i] * r;elsesuf[i] = max(suf[i + 1], num[i] * r);}LL maxn = -1;for(int i = 1; i <= n; ++i) {if(i == 1) {maxn = pre[i] + suf[i] + num[i] * q;} else {maxn = max(maxn, pre[i] + suf[i] + num[i] * q);}}printf("%lld\n", maxn);return 0;}

C

题意：给出一颗n个结点的树，n - 1条边，每个结点有一个数字，范围在1～m，现在给定一个数字k，x，树中最多存在x个k，每个数字是k的结点周围的数字都得小于k，问有多少种满足方案

参考大神：blackcat

#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)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e5 + 10;const LL INF = 1e9 + 10;LL dp[qq][12][3];int n, m, k, x;vector<int> G[qq];void Dfs(int u, int fa) {dp[u][0][0] = k - 1, dp[u][1][1] = 1, dp[u][0][2] = m - k;int sz = G[u].size();for(int i = 0; i < sz; ++i) {int v = G[u][i];if(v == fa)continue;Dfs(v, u);LL tmp[12][3] = {0};for(int j = 0; j <= x; ++j) {for(int l = 0; l <= j; ++l) {tmp[j][0] += (dp[v][l][0] + dp[v][l][1] + dp[v][l][2]) * dp[u][j - l][0];tmp[j][0] %= MOD;tmp[j][1] += dp[v][l][0] * dp[u][j - l][1];tmp[j][1] %= MOD;tmp[j][2] += (dp[v][l][0] + dp[v][l][2]) * dp[u][j - l][2];tmp[j][2] %= MOD;}}memcpy(dp[u], tmp, sizeof tmp);}}int main(){scanf("%d%d", &n, &m);for(int a, b, i = 1; i < n; ++i) {scanf("%d%d", &a, &b);G[a].pb(b), G[b].pb(a);}scanf("%d%d", &k, &x);Dfs(1, -1);LL ans = 0;for(int i = 0; i <= x; ++i) {ans += (dp[1][i][0] + dp[1][i][1] + dp[1][i][2]);ans %= MOD;}printf("%lld\n", ans);return 0;}