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
salvatore13:
20160424 11:44:25
can anybody give some extreme cases that i should check? because I keep getting WA, but the general concept works well... 

Dushyant Singh:
20160421 21:23:25
@geekyadity> I think there is no problem with climits. Problem is because of overflow. When you add something to INT_MAX then it will definitely overflow.


geekyadity:
20160413 17:26:17
Use some MAX Macro like 999999999. Don't rely on climits. Took me a lot of time to figure out 

hash7:
20160403 14:46:29
took 5 w.a to learn a new concept . :D too much happy


Rakend Chauhan:
20160321 00:37:09
dp array locally gives RE why? 

Daniel Cárdenas:
20160316 08:04:08
Taking the longest time to solve this on my own. Last edit: 20160316 08:04:53 

killjee_15:
20160302 07:26:45
3 RE because i was declaring DP array locally :'( 

aanand:
20160210 19:01:19
nice1 problem ! 1D dp(bottom up approach) :) Last edit: 20160210 19:02:37 

Rajat Jain:
20160124 20:31:34
AC....dp!!!!!!!!! good knapsack problem.... try ur own. 

CounterNormalize:
20160124 04:57:44
Basically its coin change problem. 
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 