Quotient Polynomial

A polynomial of degree n can be expressed as

If k is any integer then we can write:

Here q(x) is called the quotient polynomial of p(x) of degree (n-1) and r is any integer which is called the remainder.

For example, if p(x) = x³ - 7x²+ 15x - 8 and k = 3 then q(x) = x² - 4x + 3 and r = 1. Again if p(x) = x³ - 7x²+ 15x - 9 and k = 3 then q(x) = x² - 4x + 3 and r = 0.

In this problem you have to find the quotient polynomial q(x) and the remainder r. All the input and output data will fit in 32-bit signed integer.

Input
Your program should accept an even number of lines of text. Each pair of line will represent one test case. The first line will contain an integer value for k. The second line will contain a list of integers (a_n, a_{n-1, …} a₀), which represent the set of co-efficient of a polynomialp(x). Here 1 ≤ n ≤ 10000. Input is terminated by <EOF>.

Output
For each pair of lines, your program should print exactly two lines. The first line should contain the coefficients of the quotient polynomial. Print the reminder in second line. There is a blank space before and after the ‘=’ sign. Print a blank line after the output of each test case. For exact format, follow the given sample.

Sample Input

Output for Sample Input

3
1 –7 15 –8
3
1 –7 15 –9

q(x): 1 -4 3
r = 1

q(x): 1 -4 3
r = 0

这个题其实并不难 一开始的那次错误是我想复杂了  其实只要想通了算法很简单 开始的时候扯得太远了 水题一个   上代码

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<ctype.h>
using namespace std;
const int maxn= 10001;
int main()
{
    char ch;
    int a[maxn],n,k,r,b[maxn];
    while(scanf("%d",&k)!=EOF)
    {
        int i=1,flag=1;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        while (scanf("%d%c",&a[i],&ch))  //不要直接输入整串字符串 那样往外取数太麻烦了
        {
            if (ch=='\n') break;
            i++;
        }
        n=i;
        b[n-1]=a[1];
        for(i=n-2; i>=0; i--)
        {
            b[i]=k*b[i+1]+a[n-i];
        }
        printf("q(x):");
        for (i=n-1; i>0; i--)
            printf(" %d",b[i]);
        printf("\n");
        printf("r = %d\n",b[0]);
        printf("\n");
    }
    return 0;
}