1.题目编号：1018
2.简单题意：ACM做事之前都需要做一个预算，主要的收入是有限制的，这种想法的背后是很简单的，无论什么时候，只要ACM成员有一些小钱就存放到存钱罐里，并且这个过程是不可逆的，除非把存钱罐给打破。在足够的时间之后，存钱罐里应该有足够的钱去支付所有的东西，但是这其中有一个很大的问题，就是不可能确定里面有多少钱，有可能我们会打破存钱罐当里面还没有足够的钱，很明显，我们想打破这个尴尬的情况，最可能的是称重来确定里面的钱数，假设我们能够确定存钱罐的准确重量，所给各类钱币的重量，然后我们就可以确定存钱罐里面最小的数量，给T组数据，每组数据给出存钱罐的重量e和装满钱币的重量f，然后给硬币种类的数量n，及每种硬币的价值及重量p、w。如果可以装满就输出最小的钱数，否则输出不可能。
3.解题思路形成过程：今天老师刚讲了动态规划的三种背包问题，这个题属于完全背包问题，而且老师上课也讲这个题了，只是平时都是求最大值，，现在求的是最小值。
4.感悟：貌似这种题就是套公式=.=
5.AC的代码：
#include<iostream>
#include<cstring>
using namespace std;
int m=10000000;
int main(){
//freopen("1.txt","r",stdin);
int T,e,f,n,p,w,w1;
int dp[10000];
cin>>T;
while(T--){
cin>>e>>f;
w1=f-e;
cin>>n;
for (int i=1;i<=w1;i++)
dp[i]=m;
dp[0]=0;
for (int j=0;j<n;j++){
cin>>p>>w;
for (int v=w;v<=w1;v++)
if (dp[v]>dp[v-w]+p){
dp[v]=dp[v-w]+p;
}}
if (dp[w1]!=m){
cout<<"The minimum amount of money in the piggy-bank is "<<dp[w1]<<"."<<endl;
}
else
cout<<"This is impossible."<<endl;
}
return 0;
}
原题：
Problem 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. <br><br>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! <br>
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. <br>
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.". <br>
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.
