Description

題庫鏈接

給你兩個正整數 \(p,k\) ，詢問是否能夠構造多項式 \(f(x)=\sum\limits_{i=0}^{d-1}a_ix^i\) ，使得存在多項式 \(q(x)\) ，滿足 \(f(x)=q(x)\cdot(x+k)+p\) 。且 \(a_i\in[0,k),i\in [0,d)\) 。

\(1\leq p\leq 10^{18},2\leq k\leq 2000\)

Solution

我們假設 \(q(x)=\sum\limits_{i=0}^{d-2}b_ix^i\) ，那麽存在 \[\begin{aligned}a_0&=kb_0+p\\a_1&=kb_1+b_0\\&\vdots\\a_{d-2}&=kb_{d-2}+b_{d-3}\\a_{d-1}&=b_{d-2}\end{aligned}\]

逐步從下往上遞推，最終我們可以得到 \(p=\sum\limits_{i=0}^{d-1} (-k)^ia_i\) 。顯然 \(p_{(10)}=\overline{a_{d-1}\cdots a_1a_0}_{(-k)}\) ，做一遍進制轉換就好了。

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('\n'))
#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;
}

[Codeforces 933B]A Determined Cleanup