Codeforces 360C Levko and Strings dp

来源：互联网发布：贵州师范大学网络平台编辑：程序博客网时间：2024/04/27 19:05

题目链接：点击打开链接

题意：

给定长度为n的字符串s，常数k

显然s的子串一共有 n(n-1)/2 个

要求找到一个长度为n的字符串t，使得t对应位置的k个子串字典序>s

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>#include<vector>#include<set>using namespace std;#define N 2505#define mod 1000000007#define ll __int64ll n,k;ll dp[N][N];//dp[i][j]表示i位置产生的j对（1-i-1都是相同的）char s[N];ll num[N], sum[N];ll work(){if(k==0)return num[1];dp[0][0] = 1;sum[0] = 1;ll ans = 0;for(ll i = 1; i <= n; i++){ll len = n-i+1;for(ll j = 0; j <= k; j++) {for(ll z = i-1; z>=0 && (i-z)*len<=j; z--) {dp[i][j] = (dp[i][j]+dp[z][j-(i-z)*len])%mod;}dp[i][j] = dp[i][j]*('z'-s[i])%mod;}ans = (ans+dp[i][k]*num[i+1]%mod)%mod;for(ll j = 0; j <= k; j++) {dp[i][j] = (dp[i][j]+sum[j]*(s[i]-'a')%mod)%mod;sum[j] = (sum[j]+dp[i][j])%mod;}}return ans;}int main(){ll i;while(cin>>n>>k){cin>>s+1;num[n+1] = 1;for(i=n;i;i--)num[i] = num[i+1]*(s[i]-'a'+1)%mod;cout<<work()<<endl;}return 0;}

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