杭电1009-FatMouse' Trade（贪心）

青旅半醒 2022-08-08 14:53 77阅读 0赞

# FatMouse' Trade #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)**

Total Submission(s): 54937 Accepted Submission(s): 18422

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

这是个简单贪心，先计算出那种交易最值得，就先进行那个，直到自己的资金为0

AC代码：

#include<cstdio>
    #include<algorithm>
    using namespace std;
    struct node 
    {
    	int v,w;
    	double q;
    }s[10000];
    bool cmp(node x,node y)
    {
    	return x.q>y.q;
    }
    int main()
    {
    	int i,n,m;
    	double a[10000];
    	while(scanf("%d%d",&m,&n),m!=-1||n!=-1)
    	{
    		for(i=0;i<n;i++)
    		{
    			scanf("%d%d",&s[i].v,&s[i].w);
    			s[i].q=s[i].v*1.0/s[i].w;
    		}
    		sort(s,s+n,cmp);
    		double sum=0;
    		for(i=0;i<n;i++)
    		{
    			if(s[i].w<=m)
    			{
    				sum+=s[i].v;
    				m-=s[i].w;
    			}
    			else
    			{
    				sum+=m*s[i].q;
    				break;
    			}
    		}
    		printf("%.3lf\n",sum);
    	}
    	return 0;
     }