UVA - 10617 Again Palindrome

题意：只通过删除，问可以弄成多少的回文串，跟之前的一题是相似的，对于str[i] = str[j]的情况，如果我们步删除的话dp[i][j] = dp[i+1][j-1] +1,如果删除str[i]的话，dp[i][j] += dp[i+1][j],相似的还有dp[i][j] += dp[i][j-1],除去相同的dp[i+1][j-1],

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 70;

char str[MAXN];
long long dp[MAXN][MAXN];

int main(){
    int t;
    scanf("%d",&t);
    while (t--){
       scanf("%s",str);
       memset(dp,0,sizeof(dp));
       int len = strlen(str);
       for (int i = 0; i < len; i++)
           dp[i][i] = 1;
        
       for (int i = len - 2; i >= 0; i--)
           for (int j = i + 1; j < len; j++){
               if (str[i] == str[j])
                   dp[i][j] = dp[i+1][j-1] + 1;
               dp[i][j] += dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1];
           }
        printf("%lld\n",dp[0][len-1]);
    }
    return 0;
}