lightoj 1013

来源：互联网发布：河北移动crm软件编辑：程序博客网时间：2024/06/05 00:59

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\n", ++CASE, ans, dp[ans][len1][len2]);    }    return 0;}

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