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;
}

NBUT 1225 NEW RDSP MODE I

[1225] NEW RDSP MODE I