UVa 10465-Homer Simpson DP完全リュックサック

Problem C:
Homer Simpson
Time Limit: 3 seconds Memory Limit: 32 MB
Homer Simpson, a very smart guy, likes eating Krusty-burgers. It takes Homer m minutes to eat a Krusty- burger. However, there�s a new type of burger in Apu�s Kwik-e-Mart. Homer likes those too. It takes him n minutes to eat one of these burgers. Given t minutes, you have to find out the maximum number of burgers Homer can eat without wasting any time. If he must waste time, he can have beer.
Input
Input consists of several test cases. Each test case consists of three integers m, n, t (0 < m,n,t < 10000). Input is terminated by EOF.
Output
For each test case, print in a single line the maximum number of burgers Homer can eat without having beer. If homer must have beer, then also print the time he gets for drinking, separated by a single space. It is preferable that Homer drinks as little beer as possible.
Sample Input

3 5 54
3 5 55

Sample Output

18
17

Problem setter: Sadrul Habib Chowdhury Solution author: Monirul Hasan (Tomal)
Time goes, you say? Ah no! Alas, Time stays, we go. -- Austin Dobson

#include <cstdio>
#include <cstring>
const int MAX = 10000+5;
const int N = 3;

int w[N], t;
int c[MAX], number[MAX];

int max(int a, int b)
{
	return a>b?a:b;
}

void knapsack(int n)
{
	memset(c, 0, sizeof(c));
	memset(number, 0, sizeof(number));
	for(int i=1; i <= n; i++)
	{
		for(int j=w[i]; j<=t; j++)
		{
			if(c[j-w[i]]+w[i] > c[j])
			{
				c[j]=c[j-w[i]]+w[i];
				number[j]=number[j-w[i]]+1;
			}else if(c[j-w[i]]+w[i] == c[j] &&
						number[j-w[i]]+1 > number[j])
				number[j]=number[j-w[i]]+1;
		}
	}
	if(c[t]==t)
		printf("%d
", number[t]);
	else
		printf("%d %d
", number[t], t-c[t]);
}

int main()
{
//	freopen("in.txt","r",stdin);
	while(scanf("%d %d %d", &w[1], &w[2], &t)!=EOF)
	{
		knapsack(2);
	}
	return 0;
}

一晩中やったのに,これは私の気持ちをどうすることもできない.