第一题！_c++终末【问题描述】你是能看到第二题的friends呢。 —

本文链接：https://blog.youkuaiyun.com/Jrzihao/article/details/54862403

探讨了zhx需要为他的后辈准备n个不同难度的问题，并希望这些问题的难度能够以一种美丽的方式排列。美丽定义为：存在一个i使得前i个数单调递增或递减，且剩余部分也单调递增或递减。任务是计算所有可能的美丽排列数量，并输出结果对p取模后的值。

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zhx's contest

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 1000). Seek EOF EOF as the end of the file.
For each case, there are two integers n n and p p separated by a space in a line. ( 1≤n,p≤1018 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

#include <cstdio>
#include <iostream>
long long mod;
using namespace std;
int main()
{
long long n,p,ans;
long long jia(long long a,long long b) ;
long long mul(long long n,long long k);
while(cin>>n>>mod)
{
if(mod==1) cout<<0<<endl;
else if(n==1) cout<<1<<endl;
else
{
ans=mul(2,n)-2;
if(ans<0) ans+=mod;
cout<<ans<<endl;
}
}
}
long long chen(long long n,long long k)
{
long long ans=0;
while(k)
{
if(k%2==1) ans=ans%mod+n%mod;
n=(n%mod)+(n%mod);
k/=2;
}
return ans;
}
long long mul(long long n,long long k)
{
long long ans=1;
while(k)
{
if(k%2==1) ans=chen(ans,n)%mod;
n=chen((n%mod),(n%mod));
k/=2;
}
return ans;
}