HDU - 5187 - zhx's contest 附mod方式的小结 ʕ •ᴥ•ʔ-优快云博客

本文探讨了一个排列组合问题，即从1到n的所有排列中找出那些能够分成两段且每段都满足单调递增或递减的排列数量。通过数学分析得出解决此问题的有效算法，并给出了具体的实现代码。

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Problem Description

As one of the most powerful brushes, zhx is required to give his juniors  problems.
zhx thinks the  problem's difficulty is . He wants to arrange these problems in a beautiful way.
zhx defines a sequence  beautiful if there is an  that matches two rules below:
1:  are monotone decreasing or monotone increasing.
2:  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 ). Seek as the end of the file.
For each case, there are two integers and separated by a space in a line. ()

Output

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

Sample Input

2 233 3 5

Sample Output

2 1

题意：从1~n有多少种排列，使得 a1~ai 满足单调递增或者单调递减，ai~an 满足单调递增或者递减。

分析：从n个数种选出i个数剩下的数要满足单调递增或者递减或者递减的规律那么方式唯一。

ans = （C(N,0)+C(N,1)+......+C(N,N)） =2^N;但是这种情况下单调递增和单调递减算了两遍因此要减2。

ans = 2^n - 2;当n为1时，要输出1，而当p为1时要输出0；

但是n，p都是LL型的，快速幂的时候会爆LL，所以这里要用到快速乘法，快速乘法其实和快速幂差不多，就是把乘号改为加

模小结：

取模有很多种方式这里总结一下

(a + b) % p = (a % p + b % p) % p

(a - b) % p = (a % p - b % p) % p

(a * b) % p = (a % p * b % p) % p

a ^ b % p = ((a % p)^b) % p

如果取模后结果为负用 ans 举例子 ans<0 (ans+p)%p

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
ll n,p;
ll mul(ll a,ll b)
{
	ll res=0;
	while(b)
	{
		if(b&1)
		res=(res+a)%p;
		a=(a+a)%p;
		b>>=1;
	}
	return res;
}
ll power(ll a,ll b)
{
	ll ans=1;
	while(b)
	{
		if(b&1)
		{
			ans=mul(ans,a)%p;
		}
		a=mul(a,a)%p;
		b>>=1;
	}
	return ans;
}
int main()
{

	while(cin>>n>>p)
	{
		if(p==1)//先判断模多少 再判断n 
		{
			cout<<0<<endl;
		}
		else if(n==1)
		cout<<1<<endl;
		else
		{
			ll ans=power(2,n)-2;
			cout<<(ans+p)%p<<endl;
		}
		
	}
	return 0;
}