Palindrome

A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome.

As an example, by inserting 2 characters, the string "Ab3bd" can be transformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome.

Input

Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters from 'A' to 'Z', lowercase letters from 'a' to 'z' and digits from '0' to '9'. Uppercase and lowercase letters are to be considered distinct.

Output

Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

Sample Input

5
Ab3bd

Sample Output

2

题意看懂了，就是给你一个字符串的长度，然后把字符串输出来，让你看一下至少加多少个字符能形成一个回文字符串。

先求出他的反串，也就是字符串逆序储存一下，然后求正串和反串的公共子序列，再用总长度减去公共子序列的长度，就是我们的答案了。这个题的dp数组开的时候要注意下，别超了内存。

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
char a[5001],b[5001];
int dp[3][5001];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        scanf("%s",a);
        int i,j;
        for(i=0; i<n; i++)
            b[i]=a[n-i-1];
        b[i]='\0';
        memset(dp,0,sizeof(dp));
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
            {
                if(a[i]==b[j])
                    dp[(i+1)%2][j+1]=dp[i%2][j]+1;
                else dp[(i+1)%2][j+1]=max(dp[(i+1)%2][j],dp[i%2][j+1]);
            }
        printf("%d\n",n-dp[n%2][n]);
    }
    return 0;
}