Return of the Aztecs
| Problem C: | Homer Simpson |
|
Time Limit: 3 seconds Memory Limit: 32 MB |
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| 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：有个人喜欢吃汉堡，一种汉堡需要m分钟，另一种汉堡需要n分钟，给你 t 分钟，不浪费任何时间，问你最多吃几个汉堡？如果必须浪费时间，最少的剩余时间，最多的汉堡。
解决方法：用暴力算法既可以解决，只需要枚举汉堡的个数就OK
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int main(){
int m,n,t;
while(cin>>m>>n>>t){
if(m<n) swap(m,n);
int ans=0,l=t;
for(int i=0;i<=t/m;i++){
int tmp=(t-i*m)%n;
if(tmp<l){
l=tmp;
ans=i;
}
}
if(l>0) cout<<ans+(t-ans*m)/n<<" "<<l<<endl;
else cout<<ans+(t-ans*m)/n<<endl;
}
return 0;
}