杭电ACM FatMouse' Trade算法题Problem DescriptionFatMouse 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 J

杭电ACM FatMouse' Trade算法题Problem DescriptionFatMouse 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 J
杭电ACM FatMouse' Trade算法题
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 2-1 -1
Sample Output
13.333

我的代码,请问哪里有错啊,谢谢
#include
int main(){
int m,i,j,k;
float rate,sum,ji,fi,t,n;
while(scanf("%f%d",&n,&m)&&n+1||m+1){
k=3*m,sum=0;
float a[k];
for(i=0;i

杭电ACM FatMouse' Trade算法题Problem DescriptionFatMouse 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 J
#include<stdio.h>
int main(){
    int m,i,j,k;
    double rate,sum,ji,fi,t,n; // double型数据
    while(scanf("%lf%d",&n,&m)!=EOF&&n+1||m+1){// 一直读到文件为  用EOF
        k=3*m,sum=0;
        double a[k];// double型数据
        for(i=0;i<k;i=i+3){
            scanf("%lf%lf",&ji,&fi);
            rate=ji/fi;
            a[i]=rate,a[i+1]=ji,a[i+2]=fi;
}
        for(i=0;i<k-3;i=i+3)
            for(j=i+3;j<k;j=j+3)
                if(a[i]<a[j]){
                    t=a[i],a[i]=a[j],a[j]=t;
                    t=a[i+1],a[i+1]=a[j+1],a[j+1]=t;
                    t=a[i+2],a[i+2]=a[j+2],a[j+2]=t;
}
//for(i=0;i<k;i=i+3) 测试不输出
//printf("%.2f ",a[i]);
for(i=0;i<k;i=i+3){
if(a[i+2]<=n)sum+=a[i+1],n-=a[i+2];
else{ sum+=n*a[i];break;}
}
printf("%.3f\n",sum);
    }
}