> #include <stdio.h>
> short int l[5001][5001];
> char s[5001];
> int love(int i, int j)//自上而下的动态规划(备忘录法)
> {
> 	if(l[i][j]!=5555)
> 		return l[i][j];
> 	if(j<=i)
> 	{
> 		l[i][j]=0;
> 		return 0;
> 	}
> 	if(s[i]==s[j])
> 	{
> 		l[i][j]=love(i+1,j-1);	
> 		return l[i][j];
> 	}
> 	else
> 	{
> 		l[i][j]=love(i+1,j);
> 		if(love(i,j-1)<l[i][j])
> 			l[i][j]=l[i][j-1];
> 		l[i][j]++;
> 		return l[i][j];
> 	}
> }
> int main()
> {
> 	int n,i,j;
> 	scanf("%d%s",&n,s);
> 	for(i=0;i<n;i++)
> 		for(j=0;j<n;j++)
> 			l[i][j]=5555;
> 	printf("%d",love(0,n-1));
> }
> /*刻画最优子结构的解为：设l[i,j]为s[i..j]之间最少需插入的字符个数。
> 		{
> 		 0							i=j;
> l[i,j]=	 l[i+1,j-1]					s[i]=s[j]&&j>=i;
> 		 min(s[i+1,j],s[i,j-1])+1	s[i]!=s[j]&&j>=i;
>         }*/