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
avik26091998:
20180807 23:02:23
Bottom up is so clinical ..Got TLE with top down.. 

be1035016:
20180609 20:29:13
use your def of macro element as return value ,even LONG_MAX doesn't work Last edit: 20180609 20:29:57 

shivampkumar:
20180607 08:29:28
Period and hyphen cost me 7 WA..Beware 

magicarp:
20180527 02:17:15
Recursive Implementation can give TLE if not implemented carefully. It took me lots of time to get it to pass the time limit. 

caro_linda2018:
20180520 03:33:32
Last edit: 20180520 03:33:45 

gurjits909:
20180331 14:56:57
@steady_bunny tnx bdy your case worked for me 

zephyr_96:
20180322 18:04:13
Just 5 lines of dp. 

amitboss:
20180202 08:36:17
finally submitted , feeling good, try to solve by 1d memo & unbound knapsack


spojabhi:
20180126 07:34:15
read unbounded knapsack from gfg :} 

shahi9935:
20180122 10:45:15
@hsakiv220290 Third case is impossible because. There are only two weights 4 and 3 and any combination of them can't lead to total weight of 5(61) 
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 