zhx's contest

Time Limit: 1000 ms / MemoryLimit: 65536 kb

Description

As one of the most powerfulbrushes, zhx is required to give his juniors $n$ problems.
zhx thinks the $i^{th}$ problem's difficulty is $i$. He wants to arrange theseproblems in a beautiful way.
zhx defines a sequence $\{a_{i}\}$ beautiful if there is an $i$ that matchestwo rules below:
1: $a_{1} .. a_{i}$ are monotone decreasing or monotone increasing.
2: $a_{i} .. a_{n}$ are monotone decreasing or monotone increasing.
He wants you to tell him that how many permutations of problems are there ifthe sequence of the problems' difficulty is beautiful.
zhx knows that the answer may be very huge, and you only need to tell him theanswer 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 aline. ($1 \leq n,p \leq 10^{18}$)

Output

For each test case, output asingle line indicating the answer.

Sample Input

2 233

3 5

Sample Output

2

1

 

Hint

In the firstcase, both sequence {1, 2} and {2, 1} are legal.

In the secondcase, 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

 

分析：

数字1到n的全排列问题

 

限制条件为：

1: a1..ai are monotone decreasing or monotone increasing.
2: ai..an are monotone decreasing or monotoneincreasing.

考虑：

a.     全递增/全递减

b.     先增→ai→再减（ai（max）=n

把n放在任意位置，除了n剩余的数都有两种排法：ai左边或者ai右边故：2^（n—1）,其中包括了情况a的两种（n在第一个位置和n在最后一个位置  故减去后:2^(n-1)-2

 

c.   先减→ai→再增 ai（min）=1

除了ai剩余的数都有两种排法：ai左边或者ai右边故：2^（n—1），其中包括了情况a的两种，减去后:2^(n-1)-2

总：a+b+c=2+2^n-4=2^n-2 种

答案：（2^n-2 ）%p

注意特殊情况，当n=1，输出1；当p=1，输出0

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;ll mod;ll mul_mod(ll a, ll n)  //快速乘取模{ll ans = 0;while (n){if (n & 1) ans = (ans + a) % mod;a = (a + a) % mod;n /= 2;}return ans;}ll pow_mod(ll a, ll n)  //快速幂取模{ll ans = 1;while (n){if (n & 1) ans = mul_mod(ans, a);a = mul_mod(a, a);n >>= 1;}return ans;}int main(){long long n;while (~scanf("%lld %lld", &n, &mod)){if (n == 1){if (mod == 1) printf("0\n");else printf("1\n");}else printf("%lld\n", (pow_mod(2, n) - 2 + mod) % mod);}return 0;}

快速幂取模：http://blog.csdn.net/paradoxo/article/details/74936496

快速乘取模：http://blog.csdn.net/paradoxo/article/details/74936435