hdu1210:Eddy's 洗牌问题



Eddy's 洗牌问题

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3281    Accepted Submission(s): 2197

Problem Description

Eddy是个ACMer,他不仅喜欢做ACM题,而且对于纸牌也有一定的研究,他在无聊时研究发现,如果他有2N张牌，编号为1,2,3..n,n+1,..2n。这也是最初的牌的顺序。通过一次洗牌可以把牌的序列变为n+1,1,n+2,2,n+3,3,n+4,4..2n,n。那么可以证明，对于任意自然数N，都可以在经过M次洗牌后第一次重新得到初始的顺序。编程对于小于100000的自然数N，求出M的值。

 

Input

每行一个整数N

 

Output

输出与之对应的M

 

Sample Input

201

 

Sample Output

202

 

Author

Eddy

 

Source

杭电ACM省赛集训队选拔赛之热身赛

 

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 不多说了，代码如下:

#include<iostream>#include<cstdio>#include<string>#include<stack>#include<functional>#include<queue>#include<algorithm>using namespace std;const int MAXN = 200001;int f[MAXN];int main(){    int n, m;    int i;    freopen("in.txt", "r", stdin);    while (cin >> n){        for (i = 1; i <= n; i++){            f[i] = 2 * i;            f[n+i] = 2 * i - 1;        }        int x = 1;        int num = 0;        while (x != (n + 1)&& x != 2 * n){            x = f[x];            num++;        }        if(x==(n+1))cout<<num+1<<endl;elsecout<<2*num<<endl;    }    return 0;}