FatMouse' Trade

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

Author
CHEN, Yue

Source
ZJCPC2004

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#include<stdio.h>#include<string.h>int J[1002], F[1002];double JF[1002];int main(){    int j, m, n, i;    double sum, temp;    while(scanf("%d%d", &m, &n) != EOF && m != -1 && n != -1)    {        memset(J, 0, sizeof(J));        memset(F, 0, sizeof(F));        memset(JF,0, sizeof(JF));        for(i=1; i<=n; ++i)  //从1开始         {            scanf("%d%d", &J[i], &F[i]);            JF[i] = J[i]*1.0/F[i];        }        for(i=2; i<=n; ++i)        {            if(JF[i] >= JF[i-1])            {                JF[0] = JF[i];                J[0] = J[i];                F[0] = F[i];                JF[i] = JF[i-1];                J[i] = J[i-1];                F[i] = F[i-1];                                for(j=i-2; j>=0; --j)                {                    if(JF[0] >= JF[j])                    {                        JF[j+1] = JF[j];                        J[j+1] = J[j];                        F[j+1] = F[j];                                            }                    else {                        JF[j+1] = JF[0];                         J[j+1] = J[0];                         F[j+1] = F[0];                         break;                    }                }            }        }         sum = 0;        temp = m;        for(i=1; i<=n; ++i)        {            temp = temp - F[i];            if(temp >= 0) sum = sum + 1.0*J[i];            else {sum = sum + (temp + F[i])*JF[i]; break;}         }        printf("%.3f\n", sum);    }       return 0; }