/*这道题是很明显的dp题，状态方程有点不大好想，也许是我刚刚接触dp的缘故吧。dp[i][j]表示字符串s1取前i个字符s2取前j个字符时最大公共子序列的大小，这样的如果s1[i]==s2[j],dp[i][j]=d[i-1][j-1]+1;

如果s1[1]!=s2[j],dp[i][j]=max{dp[i-1][j],dp[i][j-1]};*/

#include"iostream"#include"stdio.h"#include"algorithm"#include"string.h"#include"ctype.h"#include"cmath"#include"queue"#define mx 1005#define inf 32767#define max(a,b) a>b?a:busing namespace std;int dp[mx][mx];char s1[mx],s2[mx];int main(){while(scanf("%s%s",s1+1,s2+1)!=EOF){int len1=strlen(s1+1);int len2=strlen(s2+1);int i,j;memset(dp,0,sizeof(dp));for(i=1;i<=len1;i++){for(j=1;j<=len2;j++){if(s1[i]==s2[j]){dp[i][j]=dp[i-1][j-1]+1;}elsedp[i][j]=max(dp[i][j-1],dp[i-1][j]);}}cout<
      
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