UVA 10465 Homer Simpson（贪心，在给定时间内吃的最多的汉堡数目）

 Homer Simpson

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Description

Return of the Aztecs

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

题目大意：

输出不浪费时间时吃汉堡的最大数目，否则输出汉堡在最小浪费时间情况下吃汉堡的最大数目和最小浪费时间。

解题思路：

贪心，对花费时间小的先分，然后逐渐减少此汉堡个数，增加另一个汉堡的个数。

#include<iostream>
#include<cstdio>

using namespace std;

int main(){
    int m,n,t,x1,y1,temp;
    while(scanf("%d%d%d",&m,&n,&t)!=EOF){
        int flag=0,mint=10000;
        if(m>n){
            temp=m;m=n;n=temp;
        }
        int x=t/m,y=0;
        for(;x>=0;){
            if(t==m*x+n*y){
                printf("%d\n",x+y);flag=1;break;
            }
            else if((t-m*x-n*y)%n!=0&&(t-m*x-n*y)/n==0){
                if(t-m*x-n*y<mint){
                    mint=t-m*x-n*y;
                    x1=x;y1=y;
                }
                x--;
            }
            else y++;
        }
        if(flag==0) printf("%d %d\n",x1+y1,mint);
    }
    return 0;
}