NBUT 1225 NEW RDSP MODE I

• [1225] NEW RDSP MODE I

• 时间限制: 1000 ms　内存限制: 131072 K
• 问题描写叙述

• 
  

  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.

• 输入
• 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.
• 输出
• 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.
• 例子输入

• 5 1 2
  5 2 2

• 例子输出

• 2 4
  4 3

• 提示

• 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.

• 来源

• 辽宁省赛2010

  
  
  题目意思是：输入n，m，x。刚開始有一个1……n的排列。然后定义了一种操作，是将数组中的奇数位的数字选出来，依照顺序放到数组最后面。偶数位依照顺序放到奇数位的后面。进行m次这种操作之后。输出前x个数字。

  分析：找到循环节T，利用T去约m，然后再将非常小的m拿去模拟，输出前x个.
  
  

  #include <algorithm>
  #include <iostream>
  #include <cstring>
  #include <cstdlib>
  #include <cstdio>
  #include <string>
  #include <vector>
  #include <cmath>
  #include <ctime>
  #include <queue>
  #include <stack>
  #include <set>
  #include <map>
  using namespace std;
  
  typedef long long LL;
  const int mx = 1e6 + 10;
  const int inf = 0x3f3f3f3f;
  
  int num[mx];
  int n,m,x;
  int find_t() {          //找循环周期T
      int cnt=0,cur=1;
      do{
          if(cur*2<=n)
              cur*=2;
          else
              cur=(cur-n/2)*2-1;
          cnt++;
      } while(cur!=1);
      return cnt;
  }
  int main() {
      while(~scanf("%d%d%d",&n,&m,&x)) {
          for(int i=1; i<=n; i++)
              num[i]=i;
          int T=find_t();
          m%=T;
          for(int i=1; i<=x; i++) {
              if(i!=1)
                  printf(" ");
              for(int j=1; j<=m; j++) {
                  if(num[i]*2<=n)
                      num[i]*=2;
                  else
                      num[i]=(num[i]-n/2)*2-1;
              }
              printf("%d",num[i]);
          }
          printf("
  ");
      }
      return 0;
  }