FZU1062 NBUT1225

洗牌问题
设2n张牌分别标记为1, 2, ..., n, n+1, ..., 2n，初始时这2n张牌按其标号从小到大排列。经一次洗牌后，原来的排列顺序变成n+1, 1, n+2, 2, ..., 2n, n。即前n张牌被放到偶数位置2, 4, ..., 2n，而后n张牌被放到奇数位置1, 3, ..., 2n-1。可以证明对于任何一个自然数n，经过若干次洗牌后可恢复初始状态。现在你的的任务是计算对于给定的n的值(n≤10^5)，最少需要经过多少次洗牌可恢复到初始状态。

Input
输入数据由多组数据组成。每组数据仅有一个整数，表示n的值。 

Output
对于每组数据，输出仅一行包含一个整数，即最少洗牌次数。

Sample Input

10

Sample Output

6

思路：找规律；

            找数字1的位置变换；

int main()
{
int n;
while(~scanf("%d",&n))
{
int j=2,num=1;
if(n==0)
{
printf("1\n");
continue;
}
while(j!=1)
{
if(j<=n)
j*=2;
else
j=(j-n)*2-1;
num++;
}
printf("%d\n",num);
}
return 0;
}

[1225] 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

思路：两道问题都是找规律；

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<stack>
#include<iostream>
#include<cmath>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const int inf=1e9-1;
const int maxn=1000010;

int a[maxn];
int main()
{
int n,m,x;
while(~scanf("%d%d%d",&n,&m,&x))
{
//        cout<<n<<" "<<m<<" "<<x<<endl;
if(n%2==0)
n++;

int xx=1,yy=1,k=0;
while(1)
{
if(xx*2<=n)        //规律
xx=2*xx;
else
xx=xx*2-n;
k++;

if(xx==yy)
break;
//            cout<<x<<" "<<y<<endl;
}
//        cout<<k<<endl;
//        cout<<"done1"<<endl;
m=m%k;

for(int i=1;i<=n;i++)
a[i]=i;

for(int i=1;i<=x;i++)        //m已经被k约的足够小，暴力解决；
{
for(int j=0;j<m;j++)
{
if(a[i]*2<=n)
a[i]=a[i]*2;
else
a[i]=a[i]*2-n;
}
printf("%d",a[i]);
if(i!=x)
printf(" ");
}
cout<<endl;
//        cout<<"done2"<<endl;
}
return 0;
}