[Codeforces 933B]A Determined Cleanup

Description

题库链接

给你两个正整数 (p,k) ，询问是否能够构造多项式 (f(x)=sumlimits_{i=0}^{d-1}a_ix^i) ，使得存在多项式 (q(x)) ，满足 (f(x)=q(x)cdot(x+k)+p) 。且 (a_iin[0,k),iin [0,d)) 。

(1leq pleq 10^{18},2leq kleq 2000)

Solution

我们假设 (q(x)=sumlimits_{i=0}^{d-2}b_ix^i) ，那么存在 [egin{aligned}a_0&=kb_0+p\a_1&=kb_1+b_0\&vdots\a_{d-2}&=kb_{d-2}+b_{d-3}\a_{d-1}&=b_{d-2}end{aligned}]

逐步从下往上递推，最终我们可以得到 (p=sumlimits_{i=0}^{d-1} (-k)^ia_i) 。显然 (p_{(10)}=overline{a_{d-1}cdots a_1a_0}_{(-k)}) ，做一遍进制转换就好了。

Code

//It is made by Awson on 2018.2.17
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('
'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
void read(LL &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(LL x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(LL x) {if (x < 0) putchar('-'); print(Abs(x)); }

LL p, k, a[10005], d;

void work() {
    read(p), read(k); k = -k;
    while (p) {
        a[++d] = p%k, p /= k; 
        if (a[d] < 0) a[d] = -k+a[d], p++;
    }
    writeln(d);
    for (int i = 1; i <= d; i++) write(a[i]), putchar(' ');
}
int main() {
    work(); return 0;
}