Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he asks rikka to have some practice on codeforces. Then she opens the problem B:

Given an integer K,
she needs to come up with an sequence of integers A satisfying
that the number of different continuous subsequence of A is
equal to k.

Two continuous subsequences a, b are
different if and only if one of the following conditions is satisfied:

1. The length of a is
not equal to the length of b.

2. There is at least one t that at≠bt,
where at means
the t-th
element of a and bt means
the t-th
element of b.

Unfortunately, it is too difficult for Rikka. Can you help her?

 

Input

There are at most 20 testcases，each testcase only contains a single integer K (1≤K≤109)

 

Output

For each testcase print two lines.

The first line contains one integers n (n≤min(K,105)).

The second line contains n space-separated
integer Ai (1≤Ai≤n) -
the sequence you find.

 

Sample Input

 

Sample Output

 

Author

XJZX

 

Source

2015 Multi-University Training Contest 4

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <vector>

#include <cmath>

int MAX=0x3f3f3f3f;

using namespace std;

const int INF = 0x7f7f7f;

const int MAXM = 12e4+5;

int ma = 100000;

int main()

{

    int x,y,z,k;

    while(~scanf("%d",&k))

    {

        if(k == 1)

        {

            printf("1\n1\n");

            continue;

        }

        if(k == 2)

        {

            printf("2\n1 1\n");

            continue;

        }

        int flag = 1;

        if(k <= 100000)

        {

            for(int i = 1; i < k; i++)

                printf("%d ",1);

            printf("1\n");

            continue;

        }

        for(x = 0; x<= 1e5 &&flag ; x++)

            for(y = 0; x+y<=1e5 && y <= sqrt(k+0.5) && flag; y++)

            {

                int t = k - x - y - x*y;

                if(t % (x + y + 1) == 0)

                {

                    z = t  / (x +y +1);

                    if(z < 0 || x + y + z > min(k,ma))

                        continue;

                    int n = x + y + z;

                    printf("%d\n",n);

                    for(int i = 1; i <= n; i++)

                    {

                        if(i <= x)

                            printf("1 ");

                        else

                        {

                            if(i <= x + y)

                                printf("2 ");

                            else if(i < n)

                                printf("3 ");

                            else

                                printf("3\n");

                        }

                        flag = 0;

                    }

                }

            }

    }

    return 0;

}