Heritage

Description

Your rich uncle died recently, and the heritage needs to be divided among your relatives and the church (your uncle insisted in his will that the church must get something). There are N relatives (N <= 18) that were mentioned in the will. They are sorted in descending order according to their importance (the first one is the most important). Since you are the computer scientist in the family, your relatives asked you to help them. They need help, because there are some blanks in the will left to be filled. Here is how the will looks:

Relative #1 will get 1 / ... of the whole heritage,
Relative #2 will get 1 / ... of the whole heritage,
---------------------- ...
Relative #n will get 1 / ... of the whole heritage.

The logical desire of the relatives is to fill the blanks in such way that the uncle's will is preserved (i.e the fractions are non-ascending and the church gets something) and the amount of heritage left for the church is minimized.

Input

The only line of input contains the single integer N (1 <= N <= 18).

Output

Output the numbers that the blanks need to be filled (on separate lines), so that the heritage left for the church is minimized.

Sample Input

2

Sample Output

2
3
題意：將一份財(cái)產(chǎn)分成n份，1/x1, 1/x2, 1/x3, ...,1/xn.使得x1>x2>x3>...>xn,讓教堂分得的財(cái)產(chǎn)最少. xn = x1 * x2 * ...*xn-1 + 1.
如果前k個分剩下1/c,因?yàn)?/c>1/(c+1)>1/(c+2)>...所以下一個拿1/(c+1),剩下1/(c * (c+1)).
代碼：
#include<stdio.h>
#include<stdlib.h>
#define inf 1000000000
#define maxn 100000
int len1, len2, len3, len[18];
__int64 a[maxn], b[maxn], ans[maxn], res[18][maxn];
void setAns()
{
    for (int i = 0; i < len3; i++)
    {
        ans[i] = 0;
    }
}
void copy()
{
    for (int i = 0; i < len3; i++)
    {
        a[i] = ans[i];
    }
    len1 = len3;
}
void multiply()//inf進(jìn)制乘法 
{
    int i, j;
    setAns();
    for (i = 0; i < len2; i++)
    {
        for (j = 0; j < len1; j++)
        {
            ans[i+j] += a[j] * b[i];
            ans[i+j+1] += ans[i+j] / inf;
            if (ans[i+j+1] && len3 <= i + j + 1)
            {
                len3 = i + j + 2;
            }
            else if (len3 <= i + j)
            {
                len3 = i + j + 1;
            }
            ans[i+j] %= inf;
        }
    }
}
void plus()//讓a+1 
{
    int r;
    for (len2 = 0, r = 1; len2 < len1; len2++)
    {
        r += a[len2];
        if (r >= inf)
        {
            b[len2] = r - inf;
            r = 1;
        }
        else
        {
            b[len2] = r;
            r = 0;
        }
    }
    if (r)
    {
        b[len2++] = r;
    }
}
int main()
{
    int i, j, n;
    scanf("%d", &n);
    a[0] = 2;
    len1 = 1;
    printf("2\n");
    for (i = 2; i <= n; i++)
    {
        plus();
        printf("%I64d", b[len2-1]);
        for (j = len2 - 2; j >= 0; j--)
        {
            printf("%.9I64d", b[j]);//輸出占9為，前面如果不足則補(bǔ)零 
        }printf("\n");
        if (i < n)
        {
            multiply();
            copy();
        }
    }
    //system("pause");
    return 0;
}