FatMouse
时间限制:1 秒
内存限制:128 兆
特殊判题:否
提交:1431
解决:641
- 题目描述:
-
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.
- 输入:
-
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.
- 输出:
-
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.
- 样例输入:
- 5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
- 样例输出:
- 13.333
31.500#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std; struct goods
{
int J,F;
double xinjia;
}buf[]; bool cmp(goods a,goods b)
{ return a.xinjia<b.xinjia;
} int main()
{
int m,n,j,f;
while(cin>>m)
{
cin>>n;
if(m==-&&n==-) break;
int i;
for(i=;i<n;i++)
{
cin>>j>>f;
buf[i].J=j;
buf[i].F=f;
buf[i].xinjia=(double)f/j;
}
sort(buf,buf+n,cmp); double sum=0.0;
for(i=;i<n;i++)
{
if(buf[i].F<m)
{
sum=sum+buf[i].J;
m=m-buf[i].F;
}
else
{
sum=sum+buf[i].J*(double)m/buf[i].F;
break;
}
} cout<<fixed<<setprecision()<<sum<<endl; }
return ;
} /**************************************************************
Problem: 1433
User: 2009declan
Language: C++
Result: Accepted
Time:10 ms
Memory:1536 kb
****************************************************************/
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