Description
Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.

But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!

 

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it's weight in grams. 
 

Output

Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X." where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible.". 
 

Sample Input

3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4
 

Sample Output

The minimum amount of money in the piggy-bank is 60.
The minimum amount of money in the piggy-bank is 100.
This is impossible.
 
 
就是說輸入測試資料數
輸入空的存錢罐的重量以及滿了的重量
輸入硬幣種類數t
以下t行輸入t種硬幣每個硬幣的重量以及面值
求此情況下該存錢罐至少能存的錢數，無解就輸出無解
 
和硬幣那道題有相似之處，那道題是給出幾種錢幣，每種錢幣的數量無限，算資金一定的時候的錢幣的組合數
這道題是給出幾種錢幣的價值和重量，每種錢幣的數量無限，算揹包容量一定的情況下所存的最少錢數，也可以拓展到最多錢數，把min改為max就好了
動態和揹包好像
#include <stdio.h>

#define INF  0x3f3f3f3f

#include <algorithm>

using namespace std;

int t, w[], val[], dp[], w0, w1; //w0存空存錢罐重量，w1存裝滿的存錢罐重量，dp[i]代表容量為i的時候所能存的最少錢數，t錢幣種類數

void work()

{

    for(int i = ; i <= w1 - w0; i++)

        dp[i] = INF; // 初始標記為無窮，如果計算完之後任然為無窮的話說明無解，小於無窮則輸出結果

    dp[] = ;//容量為0所能存的資金也是0

    for(int i = ; i < t; i++)

    {

        for(int j = w[i]; j <= w1 - w0; j++)

            dp[j] = min(dp[j], dp[j-w[i]] + val[i]);//算出只放i種錢幣，其中第i種錢幣放（0個——所能放的最多數量）的時候存錢罐裡所存的的最少資金

    }

}

int main()

{

    int n;

    scanf("%d", &n);

    while(n--)

    {

        scanf("%d%d", &w0, &w1);

        scanf("%d", &t);

        for(int i = ; i < t; i++)

        {

            scanf("%d%d", &val[i], &w[i]);

        }

        work();

        if(dp[w1 - w0] < INF)

            printf("The minimum amount of money in the piggy-bank is %d.\n", dp[w1 - w0]);

        else

            printf("This is impossible.\n");

    }

    return ;

}