HDU——1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 54945 Accepted Submission(s): 18426



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

Sample Output

13.333
31.500

贪心，最优装载问题
#include<iostream>
#include<algorithm>
#include<stdio.h>

using namespace std;

struct data
{
int x;
int y;
double z;
}a[10010];

bool comp(data a,data b)
{
return a.z>b.z;
}

int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
if(n==-1&&m==-1)
break;
for(int i=0;i<m;i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
a[i].z=(double)a[i].x/(double)a[i].y;
}
sort(a,a+m,comp);
double sum=0.0;
for(int i=0;i<m;i++)
{
if(a[i].y>(double)n)
{
sum+=(double)a[i].x/(double)a[i].y*(double)n;
break;
}
else
{
sum+=a[i].x;
n-=a[i].y;
}
}
printf("%.3f\n",sum);
}
}