HDOJ 1009-FatMouse' Trade【贪心】

我会带着你远行 2022-07-16 17:53 171阅读 0赞

# FatMouse' Trade #

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

Total Submission(s): 73438 Accepted Submission(s): 25218

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

题目描述

就是一只老鼠有M份猫粮，这里有N个房间（就是N行数据），每个房间里都有J【i】份老鼠想要的食物 和 诱惑猫需要的猫粮 F\[i\]，所以问你这只老鼠最多能够得到多少它想要的食物，每个房间的食物可以不拿完，他给猫的猫粮和猫需要的比例\*房间有的J\[i\]的数量就是老鼠最后可以得到的。

解题思路

算出每个房间的比值，比值越大就说明越值得啊，排序之后比较即可。

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    
    struct node{
    	int aa,bb;
    	double jun;
    }shu[10000];
    
    bool cmp(node a,node b){
    	return a.jun>b.jun;
    }
    
    int main()
    {
    	int M,N;
    	while(scanf("%d%d",&M,&N)!=EOF){
    		if(M==-1&&N==-1)
    		break;
    		for(int i=0;i<N;i++){
    			int a,b;
    			scanf("%d%d",&a,&b);
    			double d=(double)a/b;
    			shu[i].aa=a;
    			shu[i].bb=b;
    			shu[i].jun=d;
    		}
    		sort(shu,shu+N,cmp);
    		//for(int i=0;i<N;i++)
    		// printf("%lf ",s[i]);
    		int dang=0;
    		double ans=0.0;
    		for(int i=0;i<N;i++){
    			if(dang+shu[i].bb<=M){
    				ans+=shu[i].aa;
    				dang+=shu[i].bb;
    			}
    			else
    			{
    				ans+=((double)(M-dang)/shu[i].bb)*shu[i].aa;
    				break	;
    			}
    		}
    		printf("%.3lf\n",ans);
    	}
    	return 0;
    }