NBUT1225-NEW RDSP MODE I

NEW RDSP MODE I
时间限制: 1000 ms　内存限制: 131072 K
问题描述
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.
样例输入
```
5 1 2
5 2 2
```
样例输出
```
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
```

题意：有1~n这样n个有序的数，每次抽出其中奇数位的放到最后，偶数位的放在前面，操作m次，问前x个数分别为什么

解题思路：暴力操作一次，找出每个位置是从上次哪个位置来的，用这个规律就可以找出循环节，也就是操作几次能够变回一开始的样子，m对循环节进行取余，然后对x位暴力地进行剩余次数的模拟

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <functional>
#include <climits>using namespace std;#define LL long long
const int INF=0x3f3f3f3f;int f[1000009];
int n,m,x;int main()
{while(~scanf("%d%d%d",&n,&m,&x)){int cnt=1;for(int i=2;i<=n;i+=2) f[cnt++]=i;for(int i=1;i<=n;i+=2) f[cnt++]=i;int pos=f[1];int res=1;while(pos!=1){res++;pos=f[pos];}m%=res;int flag=0;for(int i=1;i<=x;i++){if(flag) printf(" ");else flag=1;int k=i;for(int j=1;j<=m;j++)k=f[k];printf("%d",k);}printf("\n");}return 0;
}

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