HDU 114 Piggy-Bank
Piggy-Bank
Time
Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 2659 Accepted Submission(s):
1314
Problem Description
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 piggy-bank. 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 piggy-bank to pay everything that needs to be paid.
But there is a big problem with piggy-banks. 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 piggy-bank 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 piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!
But there is a big problem with piggy-banks. 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 piggy-bank 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 piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. 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
piggy-bank 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.".
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 piggy-bank is 60.
The minimum amount of money in the piggy-bank is 100.
This is impossible.
Source
Recommend
Eddy
1 #include<stdio.h>
2 int inf = 99999999 ;
3 int value[505] , weigt[501] , fine[10001];
4 int t , n , empty , full , V ;
5 int main ()
6 {
7 scanf ( "%d" , &t ) ;
8 while ( t -- )
9 {
10 scanf ( "%d%d" , &empty , &full ) ;
11 scanf ( "%d" , &n ) ;
12 for ( int i = 0 ; i < n ; i ++ )
13 {
14 scanf ( "%d%d" , &value[i] , &weigt[i] ) ;
15 }
16 V = full - empty ;
17 for ( int i = 0 ; i <= V ; i ++ )
18 fine[i] = inf ;
19 fine[0] = 0 ;
20 for ( int i = 0 ; i < n ; i ++ )
21 for ( int j = weigt[i] ; j <= V ; j ++ )
22 if ( fine[j] > fine[j - weigt[i]] + value[i] )
23 fine[j] = fine[j - weigt[i]] + value[i] ;
24 if ( fine[V] == inf )
25 printf ( "This is impossible.\n" ) ;
26 else
27 printf ( "The minimum amount of money in the piggy-bank is %d.\n" , fine[V] ) ;
28 }
29 return 0 ;
30 }
