FatMouse' Trade

最新推荐文章于 2022-05-09 20:02:28 发布

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最新推荐文章于 2022-05-09 20:02:28 发布

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分类专栏： ACM. 贪心算法文章标签：贪心

本文链接：https://blog.youkuaiyun.com/gjs935219/article/details/81951163

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本文探讨了FatMouse如何通过最优策略用猫粮换取最大数量的JavaBeans。问题描述了一个包含多个房间的仓库，每个房间有不同的JavaBeans和所需的猫粮比例。通过分析不同比例并按最优顺序进行交易，文章提供了一种算法来确定FatMouse能获得的最大JavaBeans数量。

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FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 94473 Accepted Submission(s): 32890

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

题目大意：

老鼠有F[i]的猫食，去找老鼠换， F[i]换J[i]；如果猫食物不够换的话，可以用百分比，来进行交换。

#include<stdio.h>
#include<algorithm>
using namespace std;
struct mm {
	double  a,b,c;
} p[1000];
bool cmp(mm d1,mm d2) {        //按照比率大小进行排序；
	return d1.c>d2.c;
}
int main() {
	double m;           //注意类型，m为double。n为int；
	int n;
	double a,b,c;
	while(~scanf("%lf%d",&m,&n)) {
		if(m==-1&&n==-1)break;
		double sum=0;
		for(int i=0; i<n; i++) {
			scanf("%lf %lf",&a,&b);
			p[i].a=a;
			p[i].b=b;
			p[i].c=a/b;
		}
		sort(p,p+n,cmp);
		for(int i=0; i<n; i++) {
			if(p[i].b<=m) {
				sum=sum+p[i].a;
				m=m-p[i].b;

			} else {

				sum=sum+(p[i].a/p[i].b)*m;
				break;
			}
		}
		printf("%.3f\n",sum);


		//for(int i=0; i<n; i++)printf("%.2f %.2f %.2f\n",p[i].a,p[i].b,p[i].c);
	}
}