UVa 766 Sum of powers (伯努利数)

题意： 求 ，要求M尽量小。

析：这其实就是一个伯努利数，伯努利数公式如下：

伯努利数满足条件B0 = 1，并且

也有

几乎就是本题，然后只要把 n 换成 n-1，然后后面就一样了，然后最后再加上一个即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 20 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

LL lcm(LL a, LL b){
  return a * (b / gcd(a, b));
}

struct Fraction{
  LL mole;
  LL deno;
  Fraction() : mole(0), deno(1){ }
  Fraction(LL m, LL d) : mole(m), deno(d) {  sinal(); }

  void sinal(){
    if(mole < 0 && deno < 0)  mole = -mole, deno = -deno;
    else if(mole >= 0 && deno < 0)  mole = -mole, deno = -deno;
    if(deno == 0)  mole = 1;
  }

  friend Fraction operator + (const Fraction &lhs, const Fraction &rhs){
    LL l = lcm(lhs.deno, rhs.deno);
    LL m = lhs.mole * (l/lhs.deno) + rhs.mole * (l/rhs.deno);
    return Fraction(m, l);
  }

  friend Fraction operator - (const Fraction &lhs, const Fraction &rhs){
    LL l = lcm(lhs.deno, rhs.deno);
    LL m = lhs.mole * (l/lhs.deno) - rhs.mole * (l/rhs.deno);
    return Fraction(m, l);
  }

  friend Fraction operator * (const Fraction &lhs, const Fraction &rhs){
    LL m = lhs.mole * rhs.mole;
    LL d = lhs.deno * rhs.deno;
    LL g = gcd(m, d);
    return Fraction(m / g, d / g);
  }

  friend Fraction operator / (const Fraction &lhs, const Fraction &rhs){
    LL m = lhs.mole * rhs.deno;
    LL d = lhs.deno * rhs.mole;
    LL g = gcd(m, d);
    return Fraction(m / g, d / g);
  }

  void print(){
    printf("%lld / %lld
", mole, deno);
  }
};

Fraction C[maxn][maxn];
Fraction B[maxn];

void init(){
  for(int i = 0; i < 25; ++i)
    C[i][0] = C[i][i] = Fraction(1, 1);
  for(int i = 2; i < 25; ++i)
    for(int j = 1; j < i; ++j)
      C[i][j] = C[i-1][j] + C[i-1][j-1];
  B[0] = Fraction(1, 1);
  for(int i = 1; i < 23; ++i){
    for(int j = 0; j < i; ++j)
      B[i] = B[i] + C[i+1][j] * B[j];
    B[i] = B[i] * Fraction(-1LL, i+1LL);
  }
}

Fraction ans[maxn];

int main(){
  init();
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    LL l = 1;
    for(int i = 1; i <= n+1; ++i){
      ans[i] = C[n+1][i] * B[n+1-i] * Fraction(1LL, n+1LL);
      l = lcm(l, ans[i].deno);
    }
    ans[n] = ans[n] + Fraction(1LL, 1LL);
    printf("%lld", l);
    for(int i = n+1; i > 0; --i)
      printf(" %lld", l / ans[i].deno * ans[i].mole);
    printf(" 0
");
    if(T)  printf("
");
  }
  return 0;
}