LDU暑假集训（十）1209 Problem I NEW RDSP MODE I

题目描述

 

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.

题意：从1—n，n个数，操作m次之后输出前k个数，1~N个数字组成数组A，将数组中奇数位组成一个数组B，偶数位组成一个数组C，然后将B连接到C后面，求执行M次，输出最后的前X位数

思路：

分析：这道题可以理解成求一个数在M次变换之后最后的位置，那么首先就从两次变换的关系开始推导

假设本次的位置为X，分为偶数和奇数两种情况：

如果X为偶数，那么X下一次的坐标就会是X'=X/2，可以理解成前面有一半的奇数被拿走了，变换一下 X=2*X'

如果X为奇数，那么X下一次的坐标就会是X'=N/2+X/2，可以理解成前面有N/2个偶数，然后还有X/2个奇数排在X前面，变换一下 X=2*X' - N

可以发现，无论是奇数还是偶数，都是 X=2*X 之后对N取模，那么经过M次变换可以看成2的M次幂，这里写一个快速幂就行了。

还要注意，当N为偶数的时候要+1，因为N为偶数的时候，取模后下标可能为0，当N为奇数时，第N位数在变换中位置是不变的，所以情况和N-1一样。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll pow(ll a,ll b,ll mod)
{
    ll res=1;
    for(;b>0;b/=2)
    {
        if(b%2==1)
            res=(res*a)%mod;
        a=(a*a)%mod;

    }
    return res;
}
int main()
{
    ll n,m,x;
    while(scanf("%lld%lld%lld",&n,&m,&x)!=EOF)
    {
        if(n%2==0)
            n++;
        ll sum=pow(2,m,n);
        ll ans=sum;
        printf("%lld",sum);
        for(int i=2;i<=x;i++)
        {
            ans+=sum;
            ans%=n;
            printf(" %lld",ans);
        }
        printf("\n");



    }



    return 0;
}