NEW RDSP MODE I （找规律）

NEW RDSP MODE I

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 25 2 2

Sample Output

2 44 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 Oneis 2,4,1,3,5. After the second operation, the sequence One is 4,3,2,1,5.So,output 4 3.

题解：给你三个数字，N，M,X。求1至N个数字经过M次变幻之后前X个数字……

变幻：奇数位置的数放在偶数位置的数的后面……

思路：抓住一个点，模拟后面各次变幻的位置，很容易就找到规律的……

#include<cstdio>#include<iostream>#include<algorithm>#include<string.h>#include<stdlib.h>#include<time.h>#include<string>#include<math.h>#include<map>#include<queue>#include<stack>#define INF 0x3f3f3f3f#define ll __int64#define For(i,a,b) for(int i=a;i<b;i++)#define sf(a)  scanf("%d",&a)#define sfs(a)  scanf("%s",a)#define sff(a,b)  scanf("%d%d",&a,&b)#define sfff(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)#define pf(a) printf("%d\n",a)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;ll fun(ll x,ll y){    ll r=1,base=2;    while(x)    {        if(x&1)r=(r*base)%y;        base=(base*base)%y;        x/=2;    }    return r;}int main(){    ll a,b,c;    while(~scanf("%I64d%I64d%I64d",&a,&b,&c))    {        if(a%2==0)a++;        ll temp=fun(b,a);        printf("%I64d",temp);        ll num=temp;        For(i,2,c+1)        {            num+=temp;            num%=a;            printf(" %I64d",num);        }        printf("\n");    }}