Uva 10465 — Homer Simpson

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

题意：在保证不休息的情况下 输出吃掉的最多汉堡的数量，若一定要休息，则使休息时间尽量的短，

分析：两个dp一个时间 一个数量，且需要注意 在时间一样的情况下，保证数量最大

key：时间相同  有不同的使用方法 所以汉堡数目不是简单的加一；也许dp！！

代码：
#include<stdio.h>
#include<string.h>
#define N 10005
#include<algorithm>
using namespace std;
int dp
;
int w[5];
int sum
;
int main()
{
    int m,n,t;
    while(scanf("%d%d%d",&w[1],&w[2],&t)!=EOF)
    {
        memset(sum,0,sizeof(sum));
        memset(dp,0,sizeof(dp));

        for(int i=1;i<=2;i++)
        {
            for(int j=w[i];j<=t;j++)
            {
                    if(dp[j]<dp[j-w[i]]+w[i])
                    {
                        dp[j]=dp[j-w[i]]+w[i];//时间为背包容量，求最多汉堡数目；
                        sum[j]=sum[j-w[i]]+1;
                    //printf("***%d   %d  %d\n",j,dp[j],sum[j]);
                    }
                    else if(dp[j]==dp[j-w[i]]+w[i])//最大时间相等则选择最多数量
                    {
                    sum[j]=max(sum[j],sum[j-w[i]]+1);
                   // printf("####%d  %d  %d\n",j,dp[j],sum[j]);
                    }
            }
        }
        if(dp[t]==t)
        printf("%d\n",sum[t]);
        else
        printf("%d %d\n",sum[t],t-dp[t]);
    }
    return 0;
}