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.
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.
(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.".
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.
#include<bits/stdc++.h>
using namespace std;
const int INF = << ;
int main()
{
int T,E,F,N,P,W;
int dp[];
while(cin>>T)
{
while(T--)
{
dp[] = ;
scanf("%d %d",&E,&F);/*E为weight of an empty pig ，F为装满硬币的存储罐的重量*/
scanf("%d",&N); /*硬币的种类*/
for(int i = ; i <= F; i++)
dp[i] = INF;
for(int i = ; i < N; i++)
{
scanf("%d %d",&P,&W); /*P为硬币的面值 W为硬币的重量*/
for(int j = W; j <= F-E; j++)
dp[j] = min(dp[j],dp[j-W]+P);
}
if(dp[F-E] == INF ) cout<<"This is impossible."<<endl;
else cout<<"The minimum amount of money in the piggy-bank is "<<dp[F-E]<<"."<<endl;
}
} return ;
}