lightoj 1013

思路：动态规划。设dp[i][j][k]表示用第一个串的前i隔字符和第二个串的前k隔字符组成长度为i的串的个数，那么:若s1[j+1] == s2[k+1] dp[i+1][j+1][k+1] += dp[i][j][k],否则：dp[i+1][j+1][k] += dp[i][j][k]; dp[i+1][j][k+1] += dp[i][j][k]

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 31;
long long int dp[MAXN<<1][MAXN][MAXN];
char s1[MAXN], s2[MAXN];
int main(){
    int t, CASE(0);
    scanf("%d", &t);
    while(t--){
        scanf("%s%s", s1, s2);
        int len1 = strlen(s1), len2 = strlen(s2);
        memset(dp, 0, sizeof dp);
        dp[0][0][0] = 1;
        for(int i = 1;i <= len1+len2;i ++){
            for(int j = 1;j <= len1+1;j ++){
                for(int k = 1;k <= len2+1;k ++){
                    if(s1[j-1] == s2[k-1]) dp[i][j][k] += dp[i-1][j-1][k-1];
                    else{
                        dp[i][j][k-1] += dp[i-1][j-1][k-1];
                        dp[i][j-1][k] += dp[i-1][j-1][k-1];
                    }
                }
            }
        }
        int ans;
        for(int i = 1;i <= len1+len2;i ++){
            if(dp[i][len1][len2]){
                ans = i;
                break;
            }
        }
        printf("Case %d: %d %lld
", ++CASE, ans, dp[ans][len1][len2]);
    }
    return 0;
}