区间DP HDU4632 容斥原理

比较新奇的区间DP题目
涉及容斥原理
题目地址

//MADE BY Y_is_sunshine;
//#include <bits/stdc++.h>
//#include <memory.h>
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>

#define INF 0x3f3f3f3f
#define MAXN 1005

const int mod = 10007;
const double PI = acos(-1);

using namespace std;

int dp[MAXN][MAXN];

int main()
{
	freopen("data.txt", "r", stdin);

	int T;
	cin >> T;
	int cnt = 0;
	while (T--) {
		string s1;
		cin >> s1;
		memset(dp, 0, sizeof(dp));
		int N = s1.size();
		s1.insert(s1.begin(), ' ');

		/*for (int i = 1; i <= N; i++) dp[i][i] = 1;*/

		for (int len = 1; len <= N; len++) {
			for (int i = 1; i + len - 1 <= N; i++) {
				dp[i][i] = 1;
				int j = i + len - 1;
				dp[i][j] = (dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1] + mod) % mod;		//容斥原理
				if (s1[i] == s1[j])
					dp[i][j] = (dp[i][j] + dp[i + 1][j - 1] + 1) % mod;							//回文 + 1
			}
		}
		printf("Case %d: ", ++cnt);
		cout << dp[1][N] << endl;

	}

	freopen("CON", "r", stdin);
	system("pause");
	return 0;
}