1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47416    Accepted Submission(s): 15987

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

 

Recommend

JGShining   |   We have carefully selected several similar problems for you:  1008 1050 1005 1051 1004 

思路：首先求出J\F的值，根据JF大小排序，有技巧J，F，JF，是一体的，所以采用冒泡，要换三个都换。然后在JF从大到小往下取。

详解代码：

#include<iostream>#include<cstdio>using namespace std;int J[1005],F[1005];double JF[1005];int main() {    int i,j;    int m,n;    while(scanf("%d%d",&m,&n)!=EOF)    {        if(m==-1&&n==-1)exit(0);        else      {        for(i=1;i<=n;i++)        {            scanf("%d",&J[i]);            scanf("%d",&F[i]);            JF[i]=J[i]*0.1/F[i]*10;        }        for(i=1;i<=n;i++)         for(j=1;j<=n-i;j++)           if(JF[j+1]>JF[j])            {                            JF[0]=JF[j+1];                J[0]=J[j+1];                F[0]=F[j+1];                                JF[j+1]=JF[j];                J[j+1]=J[j];                F[j+1]=F[j];                                JF[j]=JF[0];                J[j]=J[0];                F[j]=F[0];            }                    double sum=0;            int coun=m;            int m1;                                for(i=1;i<=n;i++)            {                m=m-F[i];                                if(m>=0)                {                    sum+=J[i];                                }                else                {                m1=m+F[i];                 sum=sum+JF[i]*m1;                               break;                }            }            printf("%0.3lf\n",sum);        }    }    return 0;    }