题干：

2018百度之星复赛晋级名单出炉（增加20%晋级名额）~

zhx's contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3779    Accepted Submission(s): 1226

 

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

 

 

Source

BestCoder Round #33

 

解题报告：

      首先这题找规律，题目推出来通式是：2^n - 2。证明如下：

推的过程就是一共有四种情况： 升升，升降，降升，降降，其中升升和降降最简单，一共有两种，复杂的就是升降和降升这两种情况，首先来看降生，那么ai一定是最小值，因为两边都算ai了，所有当在第一个空的时候，前面一共有Cn-11， 后面就自动的确定了，在第二位的时候，有Cn-12， 同理到最后Cn-1n-2，所以加起来就是2n-1-2,这是降升，同理升降也是这么多，所以最后结果就是(2n-1-2) * 2 + 2 = 2n-2;

      注意的是这题的乘法不能直接乘法，因为longlong * longlong可能会爆掉，所以这里用快速乘法，把longlong * longlong转化成longlong + longlong 去做。

AC代码：

#include<iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
typedef long long LL;

LL qmul(LL a, LL k, LL mod) { //快速乘法
	LL ans = 0;//加法的幺元
	while (k) {
		if (k & 1) ans = (ans + a)%mod;
		a = (a + a) % mod;//和快速幂一样，只不过这里是加
		k >>= 1;
	}
	return ans;
}
LL qpow(LL a, LL k, LL mod) { //快速幂
	LL ans = 1;
	while (k) {
		if (k & 1) ans = qmul(ans, a, mod);//不能直接乘
		a = qmul(a, a, mod);
		k >>= 1;
	}
	return ans;
}

int main() {
	LL n, p;
	while (~scanf("%I64d %I64d", &n, &p)) {
		if (n == 1) { //特判一下
			printf("%I64d\n", 1 % p);
			continue;
		}
		printf("%I64d\n", (qpow(2, n, p) - 2 + p) % p);//这一步注意，不要为负数
	}
	return 0;
}