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(hdu step 1.3.1)FatMouse' Trade(在收入需要一定的付出的情况下求最大收入)

2024-11-20 17:22:03   202人阅读

题目：


       

FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5092 Accepted Submission(s): 1530
 
Problem Description
FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.
 
Input
The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1‘s. All integers are not greater than 1000.
 
Output
For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.
 
Sample Input
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
 
Sample Output
13.333
31.500
 
Author
CHEN, Yue
 
Source
ZJCPC2004
 
Recommend
JGShining
 



题目大意：

       老鼠准备了M磅猫食，准备拿这些猫食跟猫交换自己喜欢的食物。有N个房间，每个房间里面都有食物。你可以得到J[i]但你需要付出F[i]的猫食。要你计算你有M磅猫食可以获得最多食物的重量。而且这里可以不必每一组都全换，可以按比例换。。。例如最后你只剩5块钱猫食。但是目前的一个选择是话10块猫食就能换取6个粮食。那么这时候你用5块猫食就能换取3个粮食



题目分析：

       这是一道简单的贪心题。对于这种获得收入的同时需要付出一些的情况下，计算最大收入。这种题一般是根据

收入和付出的比例来排一下序。然后根据这个比例从高到低进行选择




代码如下：

/*
 * a.cpp
 *
 *  Created on: 2015年1月28日
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 10005;

struct W {
	double get;//收入
	double pay;//付出的猫食
	double ave;//收入/付出比
} w[maxn];

bool cmp(W a, W b) {
	if (a.ave > b.ave) {
		return true;
	}

	return false;
}

int main() {
	int n;
	double m;
	while (scanf("%lf%d", &m, &n), n != -1 && m != -1) {
		int i;
		for (i = 0; i < n; ++i) {
			scanf("%lf %lf", &w[i].get, &w[i].pay);
			w[i].ave = w[i].get / w[i].pay;
		}

		sort(w, w + n, cmp);//贪心,对w按照get/pay进行降序排序

		double sum = 0;
		i = 0;
//		while(m >= 0){//不知道为什么这种写法就是不行.
//			if (m >= w[i].pay) {
//							sum = sum + w[i].get;
//							m = m - w[i].pay;
//						} else {
//							sum = sum + w[i].ave * m;
//							break;
//						}
//			i++;
//		}

		for (i = 0; i <= n - 1; i++) {
			if (m >= w[i].pay) {//如果当前剩余的猫食还足够的话
				sum = sum + w[i].get;//那就把那个房间的粮食全部买下
				m = m - w[i].pay;//并且手上见去相应的猫食
			} else {//如果现在手上的猫食已经不够
				sum = sum + w[i].ave * m;//那么就按比例拿去一定的猫食
				break;
			}
		}

		printf("%.3lf\n", sum);
	}

	return 0;
}



(hdu step 1.3.1)FatMouse' Trade(在收入需要一定的付出的情况下求最大收入)


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