hdoj1009解题报告

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.

题意:很久之前做的,忘了,自己看吧

#include <iostream>
#include <iomanip>
#include<cstdio>
using namespace std;

int main()
{
int m, n;
double a[10001];
int j[10001], f[10001];
while (true)
{
cin >> m >> n;
if (m == -1 && n == -1)break;
for (int i = 0; i < n; i++)
{
cin >> j[i] >> f[i];
a[i] = double(j[i]) / f[i];
}
double res = 0;
while(m != 0 && n != 0)
{
double max = -1;
int pos = 0;
for (int i = 0; i < n; i++)
if (max < a[i])
{
max = a[i];
pos = i;
}
if (m <= f[pos])
{
res = res + m * a[pos];
m = 0;
break;
}
else
{
res = res + j[pos];
m = m - f[pos];
a[pos] = 0;
}
}
printf("%.3lf\n",res);
}
return 0;
}