UVa 10719 Quotient Polynomial (mathematics)

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 are any Integ Er 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< /c11> and r = 0.

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

Input
Your program should accept a even number of lines of text. Each pair of line would represent one test case. The ' I ' would contain an integer value of K. The second line would contain a list of integers (a_n, a_{n-1, ...} a₀), which repres ENT the set of co-efficient of a polynomial p (x). Here 1≤n≤10000. The Input is terminated by <eof>.

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

Train of thought: Q (x) In addition to the maximum coefficient of the highest term and P (x) the same coefficient, the other coefficients have a law q[i] = p[i] + k * Q[i-1]. (Use large Division card)

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

Complete code:

/*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;
}