hdu_5187_zhx's contest

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

 

看了网上大神的思路：

转自：http://blog.csdn.net/qingshui23/article/details/61627248

 

首先我们将题目给定的条件在细化一下：
(1)：a1..ai是单调递增的，那么ai..an一定是单调递减的，而且 ai=n。
(2)：a1..ai是单调递减的，那么ai..an一定是单调递增的，而且 ai=1。
将条件细化成这样之后就好想多了，其实 (1) 和 (2) 是没有什么区别的，就拿第一个来说，因为 ai=n 这是确定的，所以只需要确定 i 的位置就行啦，确定好 i 的位置之后，剩下的就是简单的组合数学了，假设 n=5，那么 i 的位置有 5 个，刨除掉 1，还有 4 个数
i=1：那么符合条件的有C04个
i=2：那么符合条件的有C14个
i=3：那么符合条件的有C24个
i=4：那么符合条件的有C34个
i=5：那么符合条件的有C44个
所以总数是 24 根据二项式定理，那么满足第 (2) 个条件的也有 24 个，但是中间有重复的，就是单调递增的和单调递减的多算了以此，所以减 2，总数就是 24∗2−2，

可以推出这个题的总的方法数就是 2n−1∗2−2， 即 2n−2，还需要注意的问题就是，这个取模的数太大，所以在进行快速幂的时候不要直接乘，要进行快速乘法。

i=1与i=n的递增与递减，计算了两遍。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define maxn 100005
ll mul(ll a,ll b,ll m)
{
    ll res=0;
    while(b)
    {
        if(b&1) res+=a;
        if(res>m) res-=m;
        a+=a;
        if(a>m)
        a-=m;
        b>>=1;
    }
    return res;
}
ll pow(ll a,ll b,ll m)
{
    ll res=1;
    while(b)
    {
        if(b&1) res=mul(res,a,m);
        a=mul(a,a,m);
        b>>=1;
    }
    return res;
}

int main()
{
    ll n,p;

    while(~scanf("%lld%lld",&n,&p))
    {
        if(n==1&&p!=1)
            {puts("1");continue;}
        else if(n==1&&p==1)
            {puts("0");continue;}
        ll ans=pow(2,n,p)-2;
        ans%=p;
        ans=(ans+p)%p;
        printf("%lld\n",ans);
    }
    return 0;
}