题意：给定一个字符串，求添加最少的字母使得该串是回文串。

思路：区间dp+记忆化搜索。dp[i][j]为区间的最小添加数，那么dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1；相等时则为dp[i+1][j-1];

code：

#include <bits/stdc++.h>
using namespace std;const int INF=0x3f3f3f3f;
const int N=1005;
char s[N];
int n,dp[N][N];int dfs(int i,int j)
{if (i>j||i==j) return dp[i][j]=0;if (dp[i][j]!=INF) return dp[i][j];int& ans=dp[i][j];if (s[i]==s[j]) ans=dfs(i+1,j-1);ans=min(ans,min(dfs(i+1,j),dfs(i,j-1))+1);return dp[i][j];
}
void print(int i,int j)
{if (i>j) return;if (i==j) {printf("%c",s[i]);return;}if (s[i]==s[j]){printf("%c",s[i]);print(i+1,j-1);printf("%c",s[i]);} else if (dp[i][j]==dp[i+1][j]+1){printf("%c",s[i]);print(i+1,j);printf("%c",s[i]);}else {printf("%c",s[j]);print(i,j-1);printf("%c",s[j]);}
}
int main()
{while (~scanf("%s",s)){int n=strlen(s);memset(dp,INF,sizeof(dp));dfs(0,n-1);printf("%d ",dp[0][n-1]);print(0,n-1);puts("");}
}