NEW RDSP MODE I （快速幂）

﹏ヽ暗。殇╰゛Y 2021-09-26 16:30 213阅读 0赞

题目：

## 问题 : 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.

题目描述：

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。

还可以找出循环节t，然后t%m求余，在计算剩下的次数就可以了。

这里用的是快速幂写的。

代码：

#include <stdio.h>
    #include <string.h>
    #include <algorithm>
    using namespace std;
    long long m,n;
    long long p(long long x,long long y,long long z)
    {
        long long t = 1;
        while(y){
            if(y&1) t = (t*x)%z;
            x = (x*x)%z;
            y>>=1;
        }
        return t;
    }
    int main()
    {   
        int i,j,k,l;
        while(scanf("%lld%lld%d",&m,&n,&k)!=EOF){
            if(m%2==0) m++;
    //      n %=m;
            long long an = p(2,n,m);
            long long ans = an;
            printf("%lld",ans);
            for(i = 2;i <= k;i++){
                ans+=an;
                ans%=m;
                printf(" %lld",ans);
            }
            printf("\n");
        }
        return 0;
    }