G - NEW RDSP MODE I

 NBUT - 1225 

Time limit   1000 ms

Memory limit 131072 kB

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 2
5 2 2

Sample Output

2 4
4 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 One
is 2,4,1,3,5. After the second operation, the sequence One is 4,3,2,1,5.So,output 4 3.

题意大概为给定一个1—n的递增数列，经过m次变换，每次变换将数列中奇数位置的数取出放在数列末尾，问变换后数列

的前x个数是多少。

这道题时间限制为一秒，暴力肯定超时，所以我们要找其中的规律，n和m都很大，x却很小，所以我们可以以x为突破口。

假定数i在数列中的位置为s，经过一次变换后数i的位置S：n为偶数时s=2*S (S<=n/2),s=2*S-n-1 (n/2<S<=n);n为奇数时s=

2*S (S<=n/2),s=2*S-n (n/2<S<=n)。通个计算可以计算出变换后数列中前x个数在变化前的位置，变化前的位置就对应其值。

#include<iostream>
#include<cstdio>
using namespace std;
int location(int n,int m,int s)
{
    int cnt=0,flag=true,S=s;
    if(n%2)
        while(flag){
//            printf("!\n");
            if(s<=n/2) s*=2;
            else s=2*s-n;
            if(S==s) flag=false;
            cnt++;
        }
    else
        while(flag){
            if(s<=n/2) s*=2;
            else s=2*s-n-1;
            if(S==s) flag=false;
            cnt++;
        }
    return cnt;
}
int gbs(int a,int b)
{
    int c,A=a,B=b;
    while(b!=0){
        c=a%b;
        a=b;
        b=c;
    }
    return A*B/a;
}
int findnum(int n,int m,int s)
{
    if(n%2)
        while(m--){
            if(s<=n/2) s*=2;
            else s=2*s-n;
        }
    else
        while(m--){
            if(s<=n/2) s*=2;
            else s=2*s-n-1;
        }
    return s;
}
int main()
{
    int n,m,x;
    while(scanf("%d %d %d",&n,&m,&x)!=-1){
        int k=1;
        for(int i=1;i<=x;i++)
            k=gbs(k,location(n,m,i));//计算周期
        m%=k;
        for(int i=1;i<=x;i++){
            printf("%d",findnum(n,m,i));
            if(i!=x)
                printf(" ");
        }
        printf("\n");
    }
    return 0;
}