快速乘+快速幂 板子 zhx's contest

As one of the most powerful brushes, zhx is required to give his juniors nn problems.
zhx thinks the ithith problem's difficulty is ii. He wants to arrange these problems in a beautiful way.
zhx defines a sequence {ai}{ai} beautiful if there is an ii that matches two rules below:
1: a1..aia1..ai are monotone decreasing or monotone increasing.
2: ai..anai..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 pp.

InputMultiply test cases(less than 10001000). Seek EOFEOF as the end of the file.
For each case, there are two integers nn and pp separated by a space in a line. (1≤n,p≤10181≤n,p≤1018)OutputFor 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
        


思路:
标志为 最大最小值 （想到了图像）
所以对于每一个标志 你需要选一些数在 左成为单调 的   右边一定可以成为单调的且方案数为1
所以答案 为 sigema (1->n-1)  C(i)(n-1) 
通过杨辉三角推出 通项公式为 2^n-2;

板子  （快速乘 板子 和快速幂板子有异曲同工之妙）

// 
#include<bits/stdc++.h> 
using namespace std; 
#define ll long long 
ll n,p; 
#define maxnn 100100 
ll ksc(ll a,ll b,ll c)
{
    a=a%c;
    ll ans=0;
    while(b)
    {
        if(b&1) ans=(ans+a)%c;
        b>>=1;
        a=(a+a)%c;
    }
    return ans;
}
ll ksm(ll a,ll b,ll c)
{
    a=a%c;
    ll ans=1;
    while(b)
    {
        if(b&1) ans=ksc(ans,a,c)%c;
        b>>=1;
        a=ksc(a,a,c)%c;
    }
    return ans;
}
int main()
{
    while(cin>>n)
    {
        cin>>p;
        printf("%lld
",(ksm(2,n,p)-2+p)%p);
    }
}