FatMouse' Trade

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<iostream>#include<iomanip>using namespace std;struct aa{int a,b;double c;}ab[1002]; int main(){int m,n,i,k,l,x,a1,b1,max=1001;double a,b,c,y,c1;while(cin>>m>>n&&m!=-1&&n!=-1){k=0;for(i=0;i<n;i++){cin>>a>>b;if (b == 0)                                  c = static_cast<double>(max);//////这句不明白  elsec=a/b; if(k==0) {ab[0].a=a;ab[0].b=b;ab[0].c=c;k=1;} else  { for(l=0;l<i;l++) if(ab[l].c>c) break;  for(x=i;x>l;x--) {ab[x].a=ab[x-1].a; ab[x].b=ab[x-1].b; ab[x].c=ab[x-1].c; } ab[l].a=a; ab[l].b=b; ab[l].c=c; }}y=0;       while(m>0&&n>0)   {   n--;   if(ab[n].b<m)   y=y+ab[n].a;   else   { y=y+ab[n].c*m;break;}   m=m-ab[n].b;   }  cout<<setiosflags(ios::fixed)<<setprecision(3)<<y<<endl;}return 0;}