hdu 5187 zhx's contest(快速幂矩阵6)

zhx's contest

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1462 Accepted Submission(s): 467



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

HintIn 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

解析：给你n个数，然后把这n个数排列，规则是先升后降或者先降后升。其实规律很简单，只要确定了极值，顺序就一定了。但要注意n,p的范围，坑了好几次啊。
一共有n个位置，每个位置2种选择，共2^n中，再减去头尾各重复的一种。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
typedef long long ll;
ll mod;
ll mul(ll a, ll b)
{
    ll ans = 0;
    while(b){
        if(b&1)
          ans = (ans+a)%mod;
        a = (a+a)%mod;
        b >>= 1;
    }
    return ans;
}
ll pow(ll a, ll b)
{
    ll ans = 1;
    while(b){
        if(b&1)
          ans = mul(ans,a);
       a = mul(a, a);
       b >>= 1;
    }
    return ans;
}
int main()
{
    ll n;
    while(scanf("%lld%lld",&n,&mod))
    {
        if(n == 1)
            printf("%lld\n",n%mod);
        else printf("%lld\n",((pow((ll)2,n)-2)+mod)%mod);
    }
    return 0;
}