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