hdu 5187 高速幂高速乘法

http://acm.hdu.edu.cn/showproblem.php?pid=5187

Problem Description

As one of the most powerful brushes, zhx is required to give his juniors n problems.

zhx thinks the ith problem's
difficulty is i.
He wants to arrange these problems in a beautiful way.

zhx defines a sequence {ai} beautiful
if there is an i that
matches two rules below:

1: a1..ai are
monotone decreasing or monotone increasing.

2: ai..an are
monotone decreasing or monotone increasing.

He wants you to tell him that how many permutations of problems are there if the sequence of the problems' difficulty is beautiful.

zhx knows that the answer may be very huge, and you only need to tell him the answer module p.

 

Input

Multiply test cases(less than 1000).
Seek EOF as
the end of the file.

For each case, there are two integers n and p separated
by a space in a line. (1≤n,p≤1018)

 

Output

For each test case, output a single line indicating the answer.

 

Sample Input

2 233
3 5

 

Sample Output

2
1

Hint

In the first case, both sequence {1, 2} and {2, 1} are legal.
In the second case, sequence {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1} are legal, so the answer is 6 mod 5 = 1

/**
hdu 5187  高速幂高速乘法
题目大意：（转）数字1~n，按某种顺序排列。且满足下列某一个条件：（1）a1~ai递增，ai~an递减（2）a1~ai递减，ai~an递增。
      问有多少种不同的排列。
解题思路：首先是所有递减或所有递增各一种；另外就是满足上列两个条件的情况了。要想满足条件（1）那就仅仅能把最大的n放在i位置，
       共同拥有C（1，n-1）+C（2。n-1）+。。。

+C（n-2，n-1）即2^(n-1)-2；条件（2）与（1）同样，所以共同拥有(2^(n-1)-2)*2+2=2^n-2.
**/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long LL;

LL n,p;

LL qui_mul(LL x,LL m)///高速乘法
{
    LL re=0;
    while(m)
    {
        if(m&1)
        {
            re=(re+x)%p;
        }
        x=(x+x)%p;
        m>>=1;
    }
    return re;
}

LL qui_pow(LL a,LL n)///高速幂
{
    LL ret=1;
    LL tem=a%p;
    while(n)
    {
        if(n%1)ret=qui_mul(ret,temp)%p;
        temp=qui_mul(temp,temp)%p;
        n>>=1;
    }
    return ret;
}

int main()
{
    while(~scanf("%I64d%I64d",&n,&p))
    {
        if(n==1)
        {
            if(p==1)
                printf("0\n");
            else
                printf("1\n");
        }
        printf("%I64d\n",(qui_mul(2,n)-2)%p);
    }
    return 0;
}