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;
}