hdu1009——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 37 24 35 220 325 1824 1515 10-1 -1

Sample Output

13.33331.500

代码

#include <stdio.h>#include <iomanip>#include <algorithm>#include <cmath>using namespace std;struct Node{    double j,f,p;} s[10000];int cmp(Node x,Node y){    return x.p>y.p;}int main(){    int n,i;    double sum,m;    while(~scanf("%lf%d",&m,&n) && (m!=-1 || n!=-1))    {        sum=0;        for(i=0; i<n; ++i)        {            scanf("%lf%lf",&s[i].j,&s[i].f);            s[i].p=s[i].j/s[i].f;        }        sort(s,s+n,cmp);        for(i=0; i<n; ++i)        {            if(m>s[i].f)            {                sum+=s[i].j;                m-=s[i].f;            }            else            {                sum+=s[i].p*m;                break;            }        }        printf("%.3lf\n",sum);    }    return 0;}

完全同样的步骤，用C++语言就答案错误，用C语言就正确，真是醉了，看来要重学C？