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;
}

UVa 10719 多项式除法的商

原文：http://blog.csdn.net/u011886588/article/details/19203331