uva 10465 (Homer Simpson) (DP)

最新推荐文章于 2021-05-29 16:15:20 发布

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本文探讨了Homer Simpson如何在限定时间内最大化食用汉堡的数量，包括标准Krusty-burger和新类型的Apu's Kwik-e-Mart汉堡。通过输入参数m、n、t，我们计算出Homer最多能吃多少个汉堡，以及必要时他可以喝多少啤酒来延长时间。代码实现采用动态规划方法，确保算法效率。

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

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace  std;

const int maxn = 10010;
int cost[2], t;
int dp[maxn];

int main(){
	while(cin >> cost[0] >> cost[1] >> t){
		memset(dp , 0 , sizeof dp);
		dp[0] = 1;
		for(int i = 0;i <= t-min(cost[0] , cost[1]);i++){
			if(dp[i] != 0){
				for(int k = 0;k < 2;k++){
					if(i+cost[k] <= t){
						dp[i+cost[k]] = max(dp[i+cost[k]] , dp[i]+1);
					}
				}
			}
		}
		if(dp[t]){
			cout << dp[t]-1 << endl;
		}else{
			for(int i = t-1;i >= 0;i--){
				if(dp[i]){
					cout << dp[i]-1 << " " << t-i << endl;
					break;
				}
			}
		}
	}
	return 0;
}