NBUT

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Little A has became fascinated with the game Dota recently, but he is not a good player. In all the modes, the rdsp Mode is popular on online, in this mode, little A always loses games if he gets strange heroes, because, the heroes are distributed randomly.

Little A wants to win the game, so he cracks the code of the rdsp mode with his talent on programming. The following description is about the rdsp mode:

There are N heroes in the game, and they all have a unique number between 1 and N. At the beginning of game, all heroes will be sorted by the number in ascending order. So, all heroes form a sequence One.

These heroes will be operated by the following stages M times:

1.Get out the heroes in odd position of sequence One to form a new sequence Two;

2.Let the remaining heroes in even position to form a new sequence Three;

3.Add the sequence Two to the back of sequence Three to form a new sequence One.

After M times' operation, the X heroes in the front of new sequence One will be chosen to be Little A's heroes. The problem for you is to tell little A the numbers of his heroes.

Input

There are several test cases.
Each case contains three integers N (1<=N<1,000,000), M (1<=M<100,000,000), X(1<=X<=20).
Proceed to the end of file.

Output

For each test case, output X integers indicate the number of heroes. There is a space between two numbers. The output of one test case occupied exactly one line.

Sample Input

5 1 25 2 2

Sample Output

2 44 3

Hint

In case two: N=5,M=2,X=2,the initial sequence One is 1,2,3,4,5.After the first operation, the sequence Oneis 2,4,1,3,5. After the second operation, the sequence One is 4,3,2,1,5.So,output 4 3.
题意：n代表n个数1-n，经过m次把奇数位放在偶数位的后面的变化，输出经过m次变换后前X位。
思路：由于m比较大我们需要求出经过k次变化又变为原数列的周期T，然后由于X比较小所以我们更新的时候只需更新前X个就行了。
代码：
#include <iostream>#include<stdio.h>#include<cstdio>#include<iostream>#include<algorithm>#include<math.h>#include<string.h>#include<map>#include<vector>#include<deque>#define ll long long#define inf 0x3f3f3f3f#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int a[1000001],n;int find_T(int x)//求周期T{    int pos=1;    int k=0;    while(1)    {        k++;        if(x%2==1)//如果是奇数的话前面有偶数x/2个，奇数n+1/2个        {            x=(x+n+1)/2;        }        else            x=x/2;//偶数的话前面只有x/2个数        if(x==1)break;    }    return k;}int main(){    int m,k;    while(~scanf("%d%d%d",&n,&m,&k))    {      int T=find_T(1);      m%=T;      for(int i=1;i<=n;i++)      {          a[i]=i;      }      for(int i=1;i<=k;i++)      {          if(i!=1)            printf(" ");          for(int j=1;j<=m;j++)          {              if(a[i]*2<=n)//放偶数              {                  a[i]=2*a[i];              }              else              {                  a[i]=(a[i]-n/2)*2-1;//放奇数              }          }          printf("%d",a[i]);      }      printf("\n");    }}

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