HDU 1009 FatMouse' Trade (贪心）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 66476    Accepted Submission(s): 22593

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

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 题意：

康康准备了 M 斤的食物, 准备跟舍长交换哲学之宝 

舍长有 N 个房间. 第 i 个房间有 J[i] 的 需要 F[i] 斤的食物. 康康可以不换完整个房间的 ,

他可以用 F[i]* a% 的食物 换 J[i]* a% 的 

现在给你一个实数 M 问你康康最多能获得多少的 

#include<string.h>#include<stdio.h>#include<algorithm>using namespace std;struct node{double a,b,c;}t[11000];bool cmp(node x,node y){if(x.c==y.c)return x.a>y.a;return x.c>y.c;}int main(){int n,m,i,j,k;while(scanf("%d%d",&n,&m)!=EOF){if(n==-1&&m==-1)break;for(i=0;i<m;i++){scanf("%lf%lf",&t[i].a,&t[i].b);t[i].c=t[i].a/t[i].b;}sort(t,t+m,cmp);double sum=0.0;for(i=0;i<m;i++){if(n>=t[i].b){sum+=t[i].a;n=n-t[i].b;}else{sum+=n*t[i].c;n=0;}}printf("%.3lf\n",sum);}return 0;}