杭电ACM1009（老鼠找猫换javabean…

 

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

代码如下：

C++语言: 高亮代码由发芽网提供

#include<stdio.h>
int main()
{
int m ,n , i , j;
double a [ 1000 ],b [ 1000 ], c [ 1000 ], t , sum = 0;
while( scanf( "%d%d" , & m , &n) &&( m !=- 1||n !=- 1))
{
  for( i = 0; i <n; i ++)
    scanf( "%lf%lf" , & a [ i ], &b [ i ]);
  for( i = 0; i <n; i ++)
    if(b [ i ] == 0 ){ sum = sum + a [ i ];b [ i ] =- 1 ;} //-1没有具体意思，只是确定不是0而已！
  for( i = 0; i <n; i ++)
    c [ i ] = a [ i ] /b [ i ];
  for( j = 0; j <n - 1; j ++)            //冒泡排序，控制n-1次循环，实现n-1次比较
    for( i = 0; i <n - 1 - j; i ++)
      if( c [ i ] < c [ i + 1 ])
    {
      t = c [ i ]; c [ i ] = c [ i + 1 ]; c [ i + 1 ] = t;
      t = a [ i ]; a [ i ] = a [ i + 1 ]; a [ i + 1 ] = t; t =b [ i ];
      b [ i ] =b [ i + 1 ];b [ i + 1 ] = t;
      }
  for( i = 0; i <n; i ++)
  {
    if(b [ i ] !=- 1)
    {
      if( m >=b [ i ]){ m = m -b [ i ]; sum = sum + a [ i ];}  //先将大比例对应的javabean先取完，然后用m存储剩下的的猫粮数
      else { sum = sum + c [ i ] * m; break ;}  //当猫粮数小于小比例的仓库的猫粮数时，计算出每份猫粮能换的javabean，再乘剩下的猫粮数，加上sum得到javabean数
    }
  }
  printf( "%.3lf \n " , sum);
  sum = 0;
}
}