HDU FatMouse' Trade

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

5 37 24 35 220 325 1824 1515 10-1 -1

Sample Output

13.33331.500

Author

CHEN, Yue

Source

ZJCPC2004

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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