HDU FatMouse' Trade

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本文详细阐述了FatMouse在利用猫粮与JavaBean进行交易时的优化算法，通过输入每间房间内JavaBean与所需猫粮的数量，算法计算出FatMouse能获取的最大数量的JavaBean。采用贪心排序策略对房间价值进行排序，并依次交易价值较高的房间，最终输出最大获取的JavaBean数量。

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FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 69 Accepted Submission(s) : 18

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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

Sample Output

13.333
31.500

Author

CHEN, Yue

Source

ZJCPC2004

AC代码：

#include <stdio.h>
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
    int m, n, i, j;
    double total, max;
    while( cin>>m>>n&&m!=-1&&n!=-1)
    {
        double a[1000] = {0}, b[1000] = {0};//a为房间有的食物量，b为要换的时候用的食物量
         double f[1000];
        total = m;
        max = 0;
        for(i = 0; i < n; i ++ )
        {
            cin>>a[i]>>b[i];
            f[i] = a[i] / b[i];//比值，找到最大的，先换最大的
        }
        for(i = 0; i < n-1; ++i)
            for(j = 0; j < n-i; j++ )
                if( f[j] < f[j+1] )//排序，按价值排序
                {
                    double temp;
                    temp = f[j];
                    f[j] = f[j+1];
                    f[j+1] = temp;
                    temp = a[j];
                    a[j] = a[j+1];
                    a[j+1] = temp;
                    temp = b[j];
                    b[j] = b[j+1];
                    b[j+1] = temp;
                }
        for(i=0; i<n&&total>=b[i]; i++) //先把价值大的换掉，直到剩余的不足以换一个房间的食物
        {
            max+=a[i];
            total-=b[i];
        }
        if(i < n)//如果还有没换的房间食物，就尽可能的换了
            max+=total/b[i]*a[i];
        printf("%.3f\n", max);
    }
    return 0;
}

学习心得：

请无视我不伦不类的输入输出混搭，WA数次我才发现原来输出iomanip下的控制符是不可以的，目前只知可用c的