poj 1118 Lining Up

#include<iostream>
using namespace std;

double c[702][2];//坐标数组
double k[702];//斜率数组

int cmp( const void* a , const void* b ) 
{           
    return *(double *)a > *(double *)b ? 1 : -1; 
}   
int main()
{
    int n,cnt,max,num;
    while(cin >> n)
    {
        if(n == 0)break;
        for(int i = 1;i <= n; ++i)
            cin >> c[i][0] >> c[i][1];
        max = 2;
        for(int i = 1;i <= n; ++i)
        {
            cnt = 0;
            num = 2;//最少有2个点满足条件
            for(int j = i+1;j <= n;++j)
            {    
                if(c[i][0] - c[j][0] != 0)
                    k[cnt++] = (c[i][1] - c[j][1])/(c[i][0] - c[j][0]);
                else            
                    k[cnt++] = 2147483647;    //若斜率不存在，则用最大值代替
            }
            qsort(k,cnt,sizeof(k[0]),cmp);    //先将斜率数组排序，方便对共线点数的判定
            for(int i = 0;i < cnt-1;++i)
                if(k[i] == k[i+1])
                {
                    ++num;
                    if(num > max)
                        max = num;
                }
                else
                    num = 2;            
        }
        cout << max <<endl;
    }
    return 0;
}