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的子串一共同拥有 n(n-1)/2 个

要求找到一个长度为n的字符串t，使得t相应位置的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 __int64
ll n,k;
ll dp[N][N];//dp[i][j]表示i位置产生的j对（1-i-1都是同样的）
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;
}