E - 贪心

贪心

题目描述

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.

Iutput

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

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思路：

贪心策略：先按照价值/代价的比值来排序，肯定是先买比值大的；

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代码

#include <cstdio> #include<algorithm> using namespace std; struct Node{ double J; double F; double bi; }result[10000]; bool cmp(Node a,Node b){ return a.bi<b.bi; } int main(){ int m,n,i; double re;//表示有m磅猫食，n个房间 while(scanf("%d%d",&m,&n)&&m!=-1&&n!=-1){ re=0; for(i=0;i<n;i++){ scanf("%lf%lf",&result[i].J,&result[i].F); //将J和F的比值录入 result[i].bi=result[i].J/result[i].F; } sort(result,result+n,cmp); for(i=n-1;i>=0;i--){ if(m>=result[i].F){ m-=result[i].F; re+=result[i].J; }else{ re+=m*result[i].bi; break; } } printf("%.3lf\n",re); } }