sicily 1011. Lenny's Lucky Lotto 动态规划

//设dp[i][j]为选取的第i个数为j的方案数

//dp[i][j] = sum(dp[i-1][k]), i-1 <= k <= j/2

//ans = sum(dp[n][k])



#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;



const int N = 15;

const int M = 2005;



 unsigned long long dp[N][M];



int main()

{

	int n, m, num = 0;

	int cases;



	cin >> cases;



	while (cases--)

	{

		cin >> n >> m;

		unsigned long long ans = 0;

		memset(dp, 0, sizeof(dp));

		for (int i = 1; i <= m; i++)

			dp[1][i] = 1;



		for (int i = 2; i <= n; i++)

			for (int j = 1; j <= m; j++)		

				for (int k = i-1; k <= j/2; k++)

					dp[i][j] += dp[i-1][k];

			



		for (int i = n; i <= m; i++)

			ans += dp[n][i];

		printf("Case %d: n = %d, m = %d, # lists = %lld\n", ++num, n, m, ans);

	}

	return 0;

}