HDU 5187 zhx's contest（防爆__int64 ）

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原创最新推荐文章于 2026-06-26 17:46:59 发布 1.4k 阅读 AI 写同款· GEO 优化 ›

施密特触发电路详细原理介绍我们想获得矩形脉冲，有种方法是可以采用整形的方法获得矩形脉冲，但前提是可以找到频率和幅度都符合要求的电压信号。我们设定矩形脉冲的几个指标。施密特电路是脉冲波形变换中一种常采用的电路，主要有以下两个特点： 1.输入电平从低电平上升过程中电路状态转换为对应的输入电平，与高电平下降过程中对应的输入转换电平不同。 2.通过电路内部的正反馈使输出电压波形的边沿变得很陡。施密特电路的结构和工作原理我们先从模电的角度来分析此电路的工作原理。此电路是通过公共发射极电阻耦合的两级正反馈放大器。 v1输入为低电平阅读详情

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;
}

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