hdu--1009

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58364    Accepted Submission(s): 19546

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

 

 解体思路：求出交换l率，按交换率从大到小的顺序交换。

代码如下：

#include<stdio.h>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;struct stu{double j;double f;double v;};stu num[1000+10];int cmp(stu a,stu b){return a.v>b.v;}int main(){int m,n;while(scanf("%d%d",&m,&n)){if(n==-1&&m==-1)break;for(int i=0;i<n;i++){scanf("%lf%lf",&num[i].j,&num[i].f);if(num[i].f==0)num[i].v=INF;   else  num[i].v=num[i].j/num[i].f;}sort(num,num+n,cmp);double res,sum;res=m;sum=0;        for(int i=0;i<n;i++){if(res>=num[i].f){        res-=num[i].f;        sum+=num[i].j;        }        else if(res<=0)break;        else {        sum+=res*num[i].v;        res=0;        }        }        printf("%.3lf\n",sum);}return 0;}