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.
Added by:  Adrian Kosowski 
Date:  20040606 
Time limit:  5s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel Pentium G860 3GHz) 
Languages:  All except: SCM chicken 
Resource:  ACM Central European Programming Contest, Prague 1999 
hide comments
:.Mohib.::
20150618 11:11:32
Enjoyed.....nice que!! 

pratika:
20150616 21:22:20
I'm getting a tle even though I used dp. Does anybody have any suggestions? 

lvn_anand:
20150616 19:45:45
Don't the sample input/output(plus FE=0) cover all kinds of cases?


Vishal:
20150610 21:26:15
missed period cost me 2 WA :( 

NEXES:
20150517 05:11:21
Woo!! AC in 1 attempt...........unbound 01 Knapsack 

Tony T.:
20150514 11:37:56
remember to check the case when F  E = 0! 

eightnoteight:
20150504 06:03:40
stupid c++, can't even handle the decision of allocating memory in stack vs heap. Last edit: 20150504 06:04:38 

Shaurya:
20150408 21:06:49
@Siu ching pong: That's what o/1 knapsack with unbounded buffer is.


Sunil:
20150331 17:09:05
oops missed the period 

Siu Ching Pong (Asuka Kenji):
20150319 11:28:32
No need to use "0/1 knapsack" algorithm, 1D array of size "F  E + 1" is sufficient to solve the problem. 