题解

先通过递推式求得伯努利数，然后用公式并将中间的$(n+1) ^ i$,变成$n ^ i$,后面再加上$n ^ k$，化进去就行了。简而言之就是由数学伯努利公式代入，通分化简就可以。

标程

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <iostream>
#include <string>
#include <ctime>

#define LOCAL
const int MAXN = 20 + 10;
const double Pi = acos(-1.0);
using namespace std;

typedef long long ll;

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a % b);
}

struct Num {
    ll a, b; // 分数，b为分母
    Num(ll x = 0, ll y = 0) { a = x; b = y; }
    void update() {
        ll tmp = gcd(a, b);
        a /= tmp;
        b /= tmp;
    }
    Num operator + (const Num &c) {
        ll fz = a * c.b + b * c.a, fm = b * c.b;
        if (fz == 0) return Num(0, 1);
        ll tmp = gcd(fz, fm);
        return Num(fz / tmp, fm / tmp);
    }
} B[MAXN], A[MAXN];

ll C[MAXN][MAXN];

void init() {
    // 预处理组合数
    int i, j;
    for (i = 0; i < MAXN; i++) C[i][0] = C[i][i] = 1;
    for (i = 2; i < MAXN; i++)
        for (j = 1; j < MAXN; j++) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
    
    // 预处理伯努利数
    B[0] = Num(1, 1);
    for (i = 1; i < MAXN; i++) {
        Num tmp = Num(0, 1), add;
        for (j = 0; j < i; j++) {
            add = B[j];
            add.a *= C[i + 1][j];
            tmp = tmp + add;
        }
        if (tmp.a) tmp.b *= -(i + 1);
        tmp.update();
        B[i] = tmp;
    }
}

void work() {
    int n;
    scanf("%d", &n);
    ll M = n + 1, flag = 0, Lcm;
    A[0] = Num(0, 1);
    
    int i;
    for (i = 1; i <= n + 1; i++) {
        if (B[n + 1 - i].a == 0) {
            A[i] = Num(0, 1);
            continue;
        }
        Num tmp = B[n + 1 - i];
        tmp.a *= C[n + 1][i]; // C[n+1][i] = C[n + 1][n + 1 - i]
        tmp.update();
        if (flag == 0) {
            Lcm = flag = tmp.b;
        }
        A[i] = tmp;
    }
    
    A[n] = A[n] + Num(n + 1, 1);
    
    for (i = 2; i <= n + 1; i++) {
        if (A[i].a == 0) continue;
        Lcm = (Lcm * A[i].b) / gcd(Lcm, A[i].b);
    }
    
    if (Lcm < 0) Lcm *= -1;
    M *= Lcm;
    
    printf("%lld", M);
    for (i = n + 1; i >= 0; i--) {
        printf(" %lld", A[i].a * Lcm / A[i].b);
    }
}

int main() {
    init();
    work();
    return 0;
}