UVa 10719 多项式除法的商

/*

* 解题思路：

* 水题、输入控制好

*/

#include <stdio.h>
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<p-1;i++ )
if( i == 0 ) num2[ i ] = num1[ i ];
else num2[ i ] = num1[ i ] + k*num2[ i-1 ];

printf("q(x):");
for( i=0;i<p-1;i++ )
printf(" %d",num2[ i ] );
puts("");
printf("r = %d\n\n",num1[ p-1 ] - num2[ p-2 ]*k*(-1));
}
return 0;
}