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.
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.
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.
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
13.333
31.500
解题思路：贪心，开一个结构体后按照平均值从大到小排序，先换掉平均值大的房间里的JavaBean，输出用printf输出，用setpricision会wa，原因在于setprecision会四舍五入丢失精度。
#include<bits/stdc++.h>
#include<stdio.h>
using namespace std;
struct jb
{
double j,f,avg;
};
bool cmp(jb x,jb y)
{
return x.avg<y.avg;
}
int main()
{
double n;
int m;
while(cin>>n>>m&&n!=-1)
{
jb s[1001];
for(int i=0;i<m;i++){cin>>s[i].j>>s[i].f;s[i].avg=s[i].j/s[i].f;}
sort(s,s+m,cmp);
double sum=0;
for(int i=m-1;i>=0;--i)
{
if(n>=s[i].f)
{
n-=s[i].f;
sum+=s[i].j;
}
else if(n<s[i].f&&n>0)
{
sum+=s[i].avg*n;
break;
}
}
printf("%.3lf\n",sum);
//cout<<fixed<<setprecision(3)<<sum<<endl;
}
return 0;
}
