hdu1009FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44925    Accepted Submission(s): 15023

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

 

#include<stdio.h>struct cat{double J;double F;double p;//单价 }a[1001];int main(){int m,n,i,j,max;double sum;cat t;while(scanf("%d%d",&m,&n)&&(m!=-1&&n!=-1)){for(i=0;i<n;i++){scanf("%lf%lf",&a[i].J,&a[i].F);a[i].p=a[i].J/a[i].F;}for(sum=0,i=0;i<n&&m>0;i++){max=i;for(j=i+1;j<n;j++){if(a[max].p<a[j].p){//a[max].p=a[j].p;//一开始错在了这一句语句上，wa了半天 max=j;//找出单价的最大值 }}t=a[i];a[i]=a[max];a[max]=t;if(m>=a[i].F){sum+=a[i].J;m-=a[i].F;}else {sum+=a[i].p*m;m=0;}}printf("%.3lf\n",sum);}return 0;}