hdu 1114 Piggy-Bank

//多重背包，刚好放满的情况
//价值达到最小的情况
//hdu 1114
#include<iostream>
using namespace std;
int v[10005], w[10005], dp[10005];
//
int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		//memset(dp, 0, sizeof(dp));//①达到最大
		memset(dp, 0x7f, sizeof(dp));//①达到最小的解法
		//F-E表示背包大小
		int E, F, n, i, j;
		scanf("%d%d%d", &E, &F, &n);
		for( i=0; i < n; i++ )
		{
			scanf("%d%d", &v[i], &w[i]);
			dp[0] = 0;
			for( j=0; j <= F-E-w[i]; j++ )
			{
				//if(dp[j]+v[i] > dp[j+w[i]])//②达到最大
				if(dp[j]+v[i] < dp[j+w[i]])//②达到最小的解法
					dp[j+w[i]] = dp[j] + v[i];	
			}
			
		}
		if(dp[F-E] == 0x7f7f7f7f)
			printf("This is impossible.\n");//放不满的输出的结果
		else
			printf("The minimum amount of money in the piggy-bank is %d.\n", dp[F-E]);
	}

	return 0;
}
/*
3
10 110
2
1 1
30 50
10 110

The minimum amount of money in the piggy-bank is 60.
*/