[codeforces934D]A Determined Cleanup

试题描述

In order to put away old things and welcome a fresh new year, a thorough cleaning of the house is a must.

Little Tommy finds an old polynomial and cleaned it up by taking it modulo another. But now he regrets doing this...

Given two integers $p$ and $k$, find a polynomial $f(x)$ with non-negative integer coefficients strictly less than $k$, whose remainder is $p$ when divided by $(x + k)$. That is, $f(x) = q(x) \cdot (x + k) + p$, where $q(x)$ is a polynomial (not necessarily with integer coefficients).

给定两个整数 $p$ 和 $k$，构造一个满足下列条件的多项式 $f(x)$：

每项系数严格小于 $k$ 且非负；
$f(x) = g(x) \cdot (x+k) + p$，其中 $g(x)$ 是个多项式，系数没有任何要求。

输入

The only line of input contains two space-separated integers $p$ and $k$ $(1 \le p \le 10^{18}, 2 \le k \le 2 000)$.

输出

If the polynomial does not exist, print a single integer $-1$, or output two lines otherwise.

In the first line print a non-negative integer $d$ — the number of coefficients in the polynomial.

In the second line print d space-separated integers $a_0, a_1, \cdots , a_{d - 1}$, describing a polynomial fulfilling the given requirements. Your output should satisfy $0 \le a_i < k$ for all $0 \le i \le d - 1$, and $a_{d - 1} \ne 0$.

If there are many possible solutions, print any of them.

输入示例1

46 2

输出示例1

7

0 1 0 0 1 1 1

输入示例2

2018 214

输出示例2

3

92 205 1

数据规模及约定

见“输入”

题解

我们假设 $f(x) = \sum_{i=0}^d a_i x^i$，然后做一下 $\frac{f(x)}{(x+k)}$ 的大除法，并将得到的 $g(x)$ 的系数写出来（假设 $g(x) = \sum_{i=0}^{d-1} b_i x^i$），会发现如下规律：

\[b_{d-1} = a_d \\
b_{d-2} = a_{d-1} - k a_d \\
b_{d-3} = a_{d-2} - k a_{d-1} + k^2 a_d \\
\cdots \\
b_0 = a_1 - k a_2 + k^2 a_3 - \cdots \\
p = a_0 - k a_1 + k^2 a_2 - \cdots = \sum_{i=0}^d (-k)^i a_i
\]

于是发现 $(a_0a_1a_2 \cdots)_{-k}$ 就是 $p$ 的 $-k$ 进制表示，上面的过程证明了它是 $p$ 的 $-k$ 进制表示是满足题目要求的必要条件；由于 $g(x)$ 没有任何约束，即 $b_i$ 可以是任意实数，充分性也显然。

负进制的转化也是同样的过程，只不过除法要做到严格的向下取整，而不是用 C++ 中默认的朝零取整。

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <algorithm>

using namespace std;

#define rep(i, s, t) for(int i = (s), mi = (t); i <= mi; i++)

#define dwn(i, s, t) for(int i = (s), mi = (t); i >= mi; i--)

#define LL long long

LL read() {

	LL x = 0, f = 1; char c = getchar();

	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }

	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }

	return x * f;

}

#define maxn 65

int cnt, A[maxn];

int main() {

	LL p = read(), k = read();

	while(p) {

		LL div = p / -k;

		if(-k * div > p) div++;

		A[cnt++] = p - (-k * div);

		p = div;

	}

	printf("%d\n", cnt);

	rep(i, 0, cnt - 1) printf("%d%c", A[i], i < cnt - 1 ? ' ' : '\n');

	return 0;

}