UVa 10719 多项式除法的商

/*

* 解题思路：

* 水题、输入控制好

*/

#include 
  
   
int main( )
{
    int i;
    int k,p;
    int flag;
    int num1[ 10005 ],num2[ 10005 ] , sum;
    char c;
    while( ~scanf("%d",&k) )
    {
        flag = p = sum = 0;
        getchar( );
        while(( c = getchar( )) != '\n' )
        {
            if( c == ' ' )
            {
                if( flag ) num1[ p++ ] = (-1)*sum;
                else num1[ p++ ] = sum;
                sum = 0;
                flag = 0;
            }
            else if( c == '-') flag = 1;
            else if( c<='9' && c>='0' ) sum = sum*10 + c-'0';
        }
        if( flag ) num1[ p++ ] = sum *( -1 );
        else num1[ p++ ] = sum;

        for( i=0;i