ZOJ 3791 An Easy Game

 思路：dp+记忆化搜索，设dp[n][m]表示s1与s2不同字符个数为n，还需要变m步的方法数，那么:

dp[n][m]  = (c[n][i]*c[N-n][K-i]) * dp[n-i+(K-i)][m-1]  (i需满足数组下标不小0)。c数组表示组合数。

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MOD 1000000009
const int MAXN = 101;
using namespace std;
long long int dp[MAXN][MAXN], c[MAXN][MAXN], N, M, K;
void init(){
    c[0][0] = 1, c[1][0] = 1, c[1][1] = 1;
    for(int i = 2; i < MAXN; i ++){
        c[i][0] = 1;
        for(int j = 1; j <= i; j ++){
            c[i][j] = c[i-1][j-1] + c[i-1][j];
            if(c[i][j] >= MOD) c[i][j] %= MOD;
        }
    }
}
int dfs(int n, int m){
    if(dp[n][m] != -1) return dp[n][m];
    if(m == 0) return dp[n][m] = (n == 0);
    dp[n][m] = 0;
    for(int i = 0; i <= K; i ++){
        if(n < i || N - n < K - i) continue;
        dp[n][m] += (((c[n][i] * c[N-n][K-i]) % MOD) * dfs(n - i + (K - i), m-1))%MOD;
        dp[n][m] %= MOD;
    }
    return dp[n][m];
}
int main(){
    string s1, s2;
    init();
    while(cin >> N >> M >> K){
        cin >> s1 >> s2;
        int cnt = 0;
        for(int i = 0; i < N; i ++)
            if(s1[i] != s2[i]) cnt ++;
        memset(dp, -1, sizeof dp);
        printf("%d\n", dfs(cnt, M));
    }
    return 0;
}