A - FatMouse' Trade

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<stdio.h>#include<algorithm>using namespace std;struct p{double a;double b;double c;};bool compare(p A,p B){return A.c>B.c;}int main(){int m,n,i,j;while(scanf("%d%d",&m,&n)!=EOF){if(m==-1&&n==-1)break;    p t[1001];        for(i=0;i<n;i++){cin>>t[i].a>>t[i].b;t[i].c=t[i].a/t[i].b;}sort(t,t+n,compare);double count=0;for(i=0;i<n;i++){count+=t[i].b;if(count>m)break;}double sum=0;//if(i==0)//printf("%.3lf\n",t[0].c*m);        if(count<m){for(j=0;j<n;j++)sum+=t[j].a;printf("%.3lf\n",sum);}else{for(j=0;j<i;j++){sum+=t[j].a;m-=t[j].b;}printf("%.3lf\n",sum+m*t[i].c);}}return 0;}

#include<iostream>#include<stdio.h>using namespace std;int main(){int m,n;int i,j;double a[1002],b[1002];double c[1002];while(scanf("%d%d",&m,&n)!=EOF){if(m==-1&&n==-1)break;for(i=0;i<n;i++){cin>>a[i]>>b[i];c[i]=a[i]/b[i];}for(i=0;i<n;i++){for(j=0;j<n-i-1;j++){if(c[j+1]>c[j]){swap(b[j],b[j+1]);swap(a[j],a[j+1]);swap(c[j],c[j+1]);}}}double count=0;for(i=0;i<n;i++){count+=b[i];if(count>m)break;}double sum=0;if(count<m){for(j=0;j<n;j++)sum+=a[j];printf("%.3lf\n",sum);}else{for(j=0;j<i;j++){sum+=a[j];m-=b[j];}printf("%.3lf\n",sum+m*c[i]);}}return 0;}