CDZSC_2015寒假新人（1）——基础 c

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 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output

13.333
31.500

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
int m,n;
int f[1200],j[1200];
double a[1200],b[1200];
int next[1000];
while((scanf("%d%d",&m,&n)&&(m!=-1||n!=-1)))
{
for(int i=0;i<n;i++)
{
scanf("%d%d",&f[i],&j[i]);
a[i]=1.0*f[i]/j[i];
}
double num=0;
for(int i=0;i<n;i++)
{
for(int k=i+1;k<n;k++)
{
if(a[i]<a[k])
{
swap(a[i],a[k]);
swap(f[i],f[k]);
swap(j[i],j[k]);
}
}
if(j[i]<m)
{
num+=f[i];
m-=j[i];
}
else
{

num+=1.0*f[i]*m/j[i];
m=0;
}
if(m==0)
{
break;
}
}
printf("%.3lf\n",num);
}
}

View Code