【数论】UVa 10586

Problem F: Polynomial Remains

Given the polynomial

　　　　　　a(x) = a_n xⁿ + ... + a₁ x + a₀,

compute the remainder r(x) when a(x) is divided by x^k+1.

The input consists of a number of cases. The first line of each case specifies the two integers n and k (0 ≤ n, k ≤ 10000). The next n+1 integers give the coefficients of a(x), starting from a₀ and ending with a_n. The input is terminated if n = k = -1.

For each case, output the coefficients of the remainder on one line, starting from the constant coefficient r₀. If the remainder is 0, print only the constant coefficient. Otherwise, print only the first d+1 coefficients for a remainder of degree d. Separate the coefficients by a single space.

You may assume that the coefficients of the remainder can be represented by 32-bit integers.

Sample Input

5 2
6 3 3 2 0 1
5 2
0 0 3 2 0 1
4 1
1 4 1 1 1
6 3
2 3 -3 4 1 0 1
1 0
5 1
0 0
7
3 5
1 2 3 4
-1 -1

Sample Output

3 2
-3 -1
-2
-1 2 -3
0
0
1 2 3 4

题意已标黄，即求出一元n次多项式除以x^k+1后的余数，若余数为0，则输出0；否则，从0次幂系数r0开始输出最后一个不为0的系数。
按照除法的模拟过程模拟即可。简单说一下模拟过程：

比如

5 2
6 3 3 2 0 1

多项式除以x^k+1后,x大于等于k的幂的系数均变为0。比如若使x^5系数变为0，则需要(x^2+1)*(6*x^3) = 6*x^5 + 6*x^3;这样就可以将原多项式的5次幂项减掉，顺带着，3次幂项系数会减6；由此规律，每抵消掉一个x次幂，其x-2次幂的系数会相应减少x次幂的系数。这样一直到使x的k次幂系数为0就好了。这样代码就很好写了。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
using namespace std;
const int N = 10005;
int n, k, c[N], i;
int main() {
    while (~scanf("%d%d", &n, &k) && n != -1 || k != -1)
    {
        for (i = 0; i <= n; i++)
            scanf("%d", &c[i]);
        for (i = n; i >= k; i--)
        {
            c[i - k] -= c[i];
            c[i] = 0;
        }
        i = n;
        while (c[i] == 0 && i >= 0) i--;
        if (i == -1) printf("0
");
        else {
            printf("%d", c[0]);
            for (int j = 1; j <= i; j++)
                printf(" %d", c[j]);
            printf("
");
        }
    }
    return 0;
}