[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;
}