操作集景

dp[i][j]， i 为 当前的字符数，j为当前字符数中选j个的字符窜的数，因为要去从则用pre记录出现重复的情况

dp[i][j] = dp[i-1][j] + dp[i-1][j-1];

dp[0][0] = 1;

dp[i][0] = 1;

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
using ll = long long;
constexpr size_t maxn = 1e3 + 5;
ll pre[maxn];
ll dp[maxn][maxn];
ll mod = 1e9 + 7;
int main()
{
	int n, k;
	string s;
	cin >> n >> k>> s;
	dp[0][0] = 1;
	s = ' ' + s;
	for(int i = 1; i <= n; ++ i){
		dp[i][0] = 1;
		for(int j = 1; j <= i; ++ j){
			dp[i][j] = dp[i-1][j] + dp[i-1][j-1];
			if(pre[s[i] - 'a'])dp[i][j] -= dp[pre[s[i] - 'a'] - 1][j-1];
			dp[i][j] %= mod;
		}
		pre[s[i] - 'a'] = i;
	}
	if(dp[n][k] < 0)
	dp[n][k] += mod;
	cout << dp[n][k] << endl;

}