2015 多校联赛 ——HDU5334（构造）

Virtual Participation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 237    Accepted Submission(s): 56
Special Judge

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

  

   10
  

 

Sample Output

  

   4
1 2 3 4
  

 

Author

XJZX

 

Source

2015 Multi-University Training Contest 4

假设用1，1...2,2....3,3....来构造，设他们的数量分别为 x    y     z

则能构成: x + y + z + x*y + y*z + x*z   （  x * z   代表三种数都包含的情况）

枚举  x  与   y  ，再判断z是否符合         （方法来源：http://blog.youkuaiyun.com/oilover/article/details/47164727）

P：  果然还是太弱，完全没想到╮(╯▽╰)╭，慢慢学，慢慢学

#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;
}