【POJ1707】【伯努利数】Sum of powers

Description

A young schoolboy would like to calculate the sum

for some fixed natural k and different natural n. He observed that calculating i^k for all i (1<=i<=n) and summing up results is a too slow way to do it, because the number of required arithmetical operations increases as n increases. Fortunately, there is another method which takes only a constant number of operations regardless of n. It is possible to show that the sum S_k(n) is equal to some polynomial of degree k+1 in the variable n with rational coefficients, i.e.,

We require that integer M be positive and as small as possible. Under this condition the entire set of such numbers (i.e. M, a_k+1, a_k, ... , a₁, a₀) will be unique for the given k. You have to write a program to find such set of coefficients to help the schoolboy make his calculations quicker.

Input

The input file contains a single integer k (1<=k<=20).

Output

Write integer numbers M, a_k+1, a_k, ... , a₁, a₀ to the output file in the given order. Numbers should be separated by one space. Remember that you should write the answer with the smallest positive M possible.

Sample Input

2

Sample Output

6 2 3 1 0

Source

Northeastern Europe 1997

【分析】

题意就是给出一个k,找一个最小的M使得中a[i]皆为整数.

这个涉及到伯努利数的一些公式，如果不知道的话基本没法做..

1. 伯努利数与自然数幂的关系：

2. 伯努利数递推式：

先通过递推式求得伯努利数，然后用1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。

/*
宋代朱敦儒
《西江月·世事短如春梦》
世事短如春梦，人情薄似秋云。不须计较苦劳心。万事原来有命。
幸遇三杯酒好，况逢一朵花新。片时欢笑且相亲。明日阴晴未定。 
*/
#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 
     for (int i = 0; i < MAXN; i++) C[i][0] = C[i][i] = 1;
     for (int i = 2; i < MAXN; i++)
     for (int 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 (int i = 1; i < MAXN; i++){
         Num tmp = Num(0, 1), add;
         for (int 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);
     for (int 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 (int 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 (int i = n + 1; i >= 0; i--) printf(" %lld", A[i].a * Lcm / A[i].b); 
}

int main(){
    
    init();
    work();
    //printf("%lld\n", C[5][3]);
    return 0;
}