PIGBANK  PiggyBank
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 piggybank. 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 piggybank to pay everything that needs to be paid.
But there is a big problem with piggybanks. 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 piggybank 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 piggybank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggybank. 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 piggybank 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.
".
Example
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 piggybank is 60. The minimum amount of money in the piggybank is 100. This is impossible.
hide comments
harshh3010:
20210320 15:14:13
Using LONG_MAX for invalid combination gave WA.. defining macro works 

kumar_anubhav:
20200816 13:33:28
what iis the meaniing of period ? 

kumar_anubhav:
20200816 02:08:50
why for the first testcase answer is 60 not 100 ? 

deba8221:
20200517 09:03:58
python gives tle what to do?


raman_111:
20200329 13:37:08
one dimensional dp first time i solved any 1D dp problem ;) 

mrmajumder:
20200319 16:05:04
Those who are looking for hints: unbounded knapsack 

rks14:
20190819 18:01:58
Kept getting WA for a freaking "." . 

jose0796:
20190811 17:01:33
Wrote "This is impossible" got WA, the point at the end "This is impossible." got a AC 

pimp_69:
20190509 08:06:43
In the second test case, it shows that min amount of money is 100 but the min amount of money is 120. With 1 coin of value(50) and 70 coins of value(1). Please look at the second test case or if there is some other method you guys are using to calculate the minimum amount then please let me know. Thank U.. 

dkkv0000:
20190211 15:35:07
1) watch tusharoy dp 01 knapsack

Added by:  adrian 
Date:  20040606 
Time limit:  5s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All 
Resource:  ACM Central European Programming Contest, Prague 1999 