NBUT 1225 NEW RDSP MODE I 2010辽宁省赛

Time limit  1000 ms

Memory limit  131072 kB

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.

Input

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.

Output

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.

Sample Input

5 1 2
5 2 2

Sample Output

2 4
4 3

Hint

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.

题意是每次操作都会把偶数位置的数提出放到最前面来，然后操作次数很大，求操作后的序列前几位

先看一下我傻逼一样的超时代码

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 using namespace std;

 struct node
 {
     int num;
     int id;
 }a[];

 bool cmp(node b,node c)
 {
     return b.id<c.id;
 }

 int main()
 {
     int n,m,k;
     int i,j;
     while(~scanf("%d%d%d",&n,&m,&k))
     {
         for(i=;i<=n;i++)
         {
             a[i].num=i;
             a[i].id=i;
         }
         //找循环结
         bool flag=true;
         for(i=;i<=m;i++)
         {
             for(j=;j<=n;j+=)
                 a[j].id/=;
             for(j=;j<=n;j+=)
                 a[j].id=(a[j].id+)/+n/;
             sort(a+,a++n,cmp);
             flag=true;
             for(j=;j<=n;j++)
             {
                 if(a[j].num!=j)
                 {
                     flag=false;
                     break;
                 }
             }
             if(flag)
                 break;
         }
         //取余
         if(m!=i-)
         {
             m%=i;
             //最后操作
             for(i=;i<=n;i++)
             {
                 a[i].num=i;
                 a[i].id=i;
             }
             for(i=;i<=m;i++)
             {
                 for(j=;j<=n;j+=)
                     a[j].id/=;
                 for(j=;j<=n;j+=)
                     a[j].id=(a[j].id+)/+n/;
                 sort(a+,a++n,cmp);
             }
             printf("%d",a[].num);
             for(i=;i<=k;i++)
                 printf(" %d",a[i].num);
             printf("\n");
         }
         else
         {
             printf("%d",a[].num);
             for(i=;i<=k;i++)
                 printf(" %d",a[i].num);
             printf("\n");
         }
     }
     return ;
 }

再看一下我们女队楼主的超强代码

 #include <iostream>
 #include<stdio.h>
 using namespace std;
 #define maxn 1000000
 int n;
 int ci;
 int a[maxn+];
 int k;
 int T(int x)
 {
     int c=;
     int cnt=;
     do{
       if(c*<=n)
       {
           c*=;
       }
       else
       {
           c=(c-n/)*-;
       }
       cnt++;
     }while(c!=);
     return cnt;
 }
 int main()
 {
     while(~scanf("%d%d%d",&n,&ci,&k))
     {
         ci%=T(n);
         for(int i=;i<=n;i++)a[i]=i;
         for(int i=;i<=k;i++)
         {

             for(int j=;j<=ci;j++)
             {
                 if(a[i]*<=n)
                 {
                     a[i]=*a[i];
                 }
                 else{
                     a[i]=(a[i]-n/)*-;
                 }
             }
             if(i==)cout<<a[i];
             else cout<<" "<<a[i];
         }
         cout<<endl;

     }
     return ;
52 }

所以我是不是个傻逼。。。

是。。