UVa 10719 Quotient Polynomial (数学)

10719 - Quotient Polynomial

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=99&page=show_problem&problem=1660

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.

InputYour 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>.

OutputFor 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.

思路：q(x)中除了最高次项的系数与p(x)最高项的系数相同，其它系数有一个规律q[i] = p[i] + k * q[i - 1].(可用大除法证之)

完整代码：

/*0.089s*/

#include<cstdio>

int p[10010], q[10010];

int main(void)
{
	int k, n;
	char c;
	while (~scanf("%d", &k))
	{
        c = 1;
		for (n = 0; c != '\n'; n++)
			scanf("%d%c", &p[n], &c);
		q[0] = p[0];
		for (int i = 1; i < n; i++)
			q[i] = p[i] + k * q[i - 1];
		printf("q(x):");
		for (int i = 0; i < n - 1; i++)
			printf(" %d", q[i]);
		printf("\nr = %d\n\n", q[n - 1]);
	}
	return 0;
}