Uva 10719 - Quotient Polynomial

Problem B	Quotient Polynomial
Time Limit	2 Seconds

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, ifp(x) = x³- 7x²+ 15x - 8andk = 3thenq(x) = x²- 4x + 3andr = 1. Again ifp(x) = x³- 7x²+ 15x - 9andk = 3thenq(x) = x²- 4x + 3and 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 polynomial p(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

 

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Problem setter:Mohammed Shamsul Alam

Special thanks to Tanveer Ahsan

 

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
int poly[10010];
int main()
{

    int r, k, cnt, i, flag;
    char temp[20], ch;
    while(scanf("%d", &k) != EOF)
    {
        getchar();
        cnt = i = flag = 0;
        while((ch = getchar()) != EOF && ch != '\n')
        {
            if(ch != ' ')
            {
                temp[i++] = ch;
                flag = 1;
            }
            else if(flag)
            {
                flag = 0;
                temp[i] = '\0';
                poly[cnt++] = atoi(temp);
                i = 0;
            }
        }
        if(flag) 
        {
            temp[i] = '\0';
            poly[cnt++] = atoi(temp);
        }
        if(cnt-1 == 0) flag = 0;
        else flag = poly[0];
        printf("q(x): %d", flag);
        for(i=1; i<cnt; ++i)
        {
            poly[i] = poly[i] + k*poly[i-1];
            if(i != cnt-1) printf(" %d", poly[i]);
        }
        printf("\nr = %d\n\n", poly[cnt-1]);
        
    }
    return 0;
}

解题思路：
题目的意思再清楚不过，给你式子中的 k，和 p(x)式子各项的系数，要你求出满足这个式子      中的q(x)的系数和 r 的值，在草稿纸上可以得出：

设 p(x) = An*x^n + An-1*x^(n-1) + An-2*x^(n-2) + ...... + A0

    q(x) = a`x^(n-1) + b`x^(n-2) + c`x^(n-3) + ...... + d`x^0

则 由 p(x) = (x-k)*q(x) + r 得：

a`x^(n-1) + b`x^(n-2) + c`x^(n-3) + ...... + d`x^0 = (x-k)*(a`x^(n-1) + b`x^(n-2) + c`x^(n-3) + ...... + d`x^0) + r

化简提起相同指数的x的指数可以得出：

a` = An;  b` = An-1 + k*a`;  c` = An-2 + k*b`;  ......  r = A0 + k*d`

这样答案就出来了

PS：不要忘记了在每一个case后都要输出一个空行，而且一开始的输入也要花点心思处理一下