题目大意：给出不同面值的金币的面值及重量，问所给重量的猪罐中最少有多少钱

输入：金币种类数T

          空猪重量E 猪罐装满后的重量F

          第i中面值的金币的价值pi  重量wi（共T行）

输出：如果刚好可以装满则输出：

The minimum amount of money in the piggy-bank is 最小钱数.

          如果不能刚好装满则输出：

This is impossible.

分析：动规。典型的完全背包问题。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int E, F,totalW;
int N;
int p[510], w[510];
int dp[10010];
int main(){
	int T;
	scanf("%d", &T);
	while (T--){
		scanf("%d %d", &E, &F);
		totalW = F - E;
		scanf("%d", &N);
		for (int i = 1; i <= N; i++){
			scanf("%d %d", &p[i], &w[i]);
		}
		memset(dp, 0x3f3f3f3f, sizeof(dp));
		dp[0] = 0;
		for (int i = 1; i <= N; i++){
			for (int j = w[i]; j <= totalW; j++){
				dp[j] = min(dp[j], dp[j - w[i]] + p[i]);
			}
		}
		if (dp[totalW] != 0x3f3f3f3f)
			printf("The minimum amount of money in the piggy-bank is %d.\n", dp[totalW]);
		else printf("This is impossible.\n");
		getchar();
	}
	getchar();
	return 0;
}