HDU FatMouse' Trade

FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 20   Accepted Submission(s) : 14

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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

按照兑换比例进行排序！然后贪心就出来了。

#include<iostream>#include<algorithm>using namespace std;struct node{double x,y;double p;}a[10100];double cmp(node a,node b){return a.p>b.p;}int main(){int n;double m;int i,j;double sum;while(true){sum=0;scanf("%lf%d",&m,&n);if(n==-1&&m==-1){break;}for(i=0;i<n;i++){scanf("%lf%lf",&a[i].x,&a[i].y);a[i].p=a[i].x/a[i].y;}sort(a,a+n,cmp);        for(i=0;i<n;i++){if(m==0){break;}else if(m>=a[i].y){m-=a[i].y;                sum+=a[i].x;}else{sum+=m*a[i].p;m=0;}}printf("%.3lf\n",sum);}return 0;}