HDU 5187 zhx's contest（防爆__int64 ）_s one of the most powerful brushes, zhx is require-优快云博客

本文探讨了在解决问题时如何通过排列组合创造出美丽而有效的序列。通过详细解释美丽序列的定义及其规则，作者展示了如何计算一系列问题难度的排列组合方式，并提供了具体的计算公式和示例输入输出，为读者提供了一种全新的思考问题解决方法。

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

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

思路：枚举减减；减加 ; 加减加加一共 2^(n-1)*2-2

1 2^(n-1) -2 2^(n-1) -2 1

//  2^n-2

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 1001


ll n,mod;

ll fdd(ll x,ll m)   //计算  x*m 居然不能直接算，否者会爆__int64 
{
	ll ans=0;
	while(m)
	{
		if(m&1) ans=(ans+x)%mod;
		x=(x+x)%mod;
		m>>=1;
	}
   return ans;
}

ll pow_(ll n,ll m)
{
     ll ans=1;

     while(m)
	 {
	 	 if(m&1) ans=fdd(ans,n);  //计算 ans*n
	 	 n=fdd(n,n);              //计算 n*n
	 	 m>>=1;
	 }
      return (ans-2+mod)%mod;
}

int main()
{
	while(~scanf("%I64d%I64d",&n,&mod))
	{
		  ll ans=2;
		  if(n==1)
		  {
            ans=n%mod;
		  	pf("%I64d\n",ans);
		  	continue;
		  }
		  printf("%I64d\n",pow_(ans,n));
	}

    return 0;
}