杭电1009

A - FatMouse' Trade

Crawling in process...Crawling failedTime Limit:1000MS Memory Limit:32768KB64bit IO Format:%I64d & %I64u

SubmitStatus

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>#include<algorithm>using namespace std;struct vis{double j,f,t;};vis ans[1100];bool cmp(vis a,vis b){    returna.t<b.t;}int main(){int n,m,i;while(scanf("%d%d",&m,&n)&&(m!=-1&&n!=-1))        误以为和while(~scanf("%d%d",&m,&n))一样，提交超时{   double s=0.0;for(i=0;i<n;i++){scanf("%lf%lf",&ans[i].j,&ans[i].f);ans[i].t=ans[i].f/ans[i].j;}sort(ans,ans+n,cmp);for(i=0;i<n;i++){if(ans[i].f<=m){s+=ans[i].j;m-=ans[i].f;}else{ s+=ans[i].j*(m/ans[i].f); break;}    }    printf("%.3lf\n",s);}return 0;}