uva 10617

当s[i] = s[j]   dp[i][j] = 1+dp[i+1][j-1]+dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1] = 1+dp[i][j-1]+dp[i+1][j]

当s[i] != s[j]   dp[i][j] = dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]

答案超int

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 100
#define INF 0x7fffffff
#define inf 10000000
#define ull unsigned long long
#define ll long long
using namespace std;

ll dp[maxn][maxn];
char s[maxn];

void solve(int n)
{
    memset(dp, 0, sizeof(dp));
    for(int i = 0; i < n; ++ i) dp[i][i] = 1;
    for(int i = n-2; i >= 0; -- i)
        for(int j = i+1; j < n; ++ j)
        {
            if(s[i] == s[j]) dp[i][j] = 1 + dp[i][j-1] + dp[i+1][j];
            else dp[i][j] = dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1];
        }
}

int main()
{
    int t, n;
    scanf("%d", &t);
    while(t --)
    {
        scanf("%s", s);
        solve(n = strlen(s));
        printf("%lld\n", dp[0][n-1]);
    }
    return 0;
}