2016 Personal Training #5 Div.2 Gym 100637J

本文探讨了在超级阶乘数制中表示任意有理数的方法，并提供了具体的算法实现细节，包括输入输出格式及示例。

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C - 寅

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

standard input/output
Statements

On the most perfect of all planets i1c5l various numeral systems are being used during programming contests. In the second division they use a superfactorial numeral system. In this system any positive integer is presented as a linear combination of numbers converse to factorials:

$\text{[math]}$

Here a₁ is non-negative integer, and integers a_k for k ≥ 2 satisfy 0 ≤ a_k < k. The nonsignificant zeros in the tail of the superfactorial number designation $\text{[math]}$ are rejected. The task is to find out how the rational number $\text{[math]}$ is presented in the superfactorial numeral system.

Input

Single line contains two space-separated integers p and q (1 ≤ p ≤ 10⁶, 1 ≤ q ≤ 10⁶).

Output

Single line should contain a sequence of space-separated integers a₁, a₂, ..., a_n, forming a number designation $\text{[math]}$ in the superfactorial numeral system. If several solution exist, output any of them.

Sample Input

Input

1 2

Output

0 1

Input

2 10

Output

0 0 1 0 4

Input

10 2

Output

唉，弄半天真是贪心，当时错两边就没再仔细看，倒是卡在数据类型半天。

题意：给你p，q输出满足公式的ai， $\text{[math]}$

思路：从a1开始，a2=p*2!/q，...;直到p为零结束。有个小疑问0<=ak<k这个条件好像没什么用？难道其中有什么...。

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define INF 0x3f3f3f3f
int main()
{
  LL p,q;
  scanf("%lld%lld",&p,&q);
  for(int i=2;;i++)
  {
    int ans=p/q;
    if(ans>=i&&i!=2)
    {
      ans=i-1;
    }
    p=(p-ans*q)*i;
    printf("%d",ans);
    if(p==0)
    {
      printf("\n");break;
    }
    printf(" ");
  }
}