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
Murat TOPAK:
20160713 13:12:48
top down in 0.20, bottom up in 0.04 in c++. 

Pragadeesh C:
20160711 00:26:18
@aexpo


aexpo:
20160708 09:46:50
If you are using a 2D matrix ,declare it globally,


baadshah_:
20160627 12:08:51
Done using top down (0.08s)


sonali9696:
20160624 23:02:48
Finally accepted in bottom up approach! TLE in top down :/ Anyone not having TLE in top down? 

saurabh789:
20160624 09:37:28
yay ac in one go!


xinnix:
20160608 18:32:25
Don't forget the full stop at the end of each case. And also use appropriate max. 

sanjay:
20160529 15:19:54
The word "exact" should make problem easier.thanks for the author.AWESOME PROBLEM :) 

praval_singhal:
20160520 14:33:25
TLE with Topdown Approach. AC with bottomup. Finally after 3 TLE Got AC :) 

vaibhavi760:
20160519 21:14:47
recursive solution TLE iterative AC 0.12(same concept) why?? someone help

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 