Hdu 5187 zhx's contest（数学）

题目链接

zhx's contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1340    Accepted Submission(s): 427

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 
    
   
   
   
   

题意：题目链接

题解：

如果 n=1 ，答案是 1 ，否则答案是 2n−2 。
证明： ai 肯定是最小的或者最大的。考虑另外的数，如果它们的位置定了的话，那么整个序列是唯一的。
那么 ai 是最小或者最大分别有 2n−1 种情况，而整个序列单调增或者单调减的情况被算了2次，所以要减2。
要注意的一点是因为 p>231 ，所以要用快速乘法。用法与快速幂相同。如果直接乘会超过long long范围，从而wa掉。

代码如下：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<stdlib.h>
#include<vector>
#define inff 0x3fffffff
#define nn 1100
#define mod 1000000007
typedef __int64 LL;
typedef unsigned __int64 LLU;
const LL inf64=inff*(LL)inff;
using namespace std;
LL n,p;
LL qiu(LL x,LL y)
{
    LL tem=x;
    LL re=0;
    while(y)
    {
        if(y%2)
        {
            re=(re+tem)%p;
        }
        y/=2;
        tem=(tem+tem)%p;
    }
    return re;
}
LL po(LL x,LL y)
{
    LL tem=x;
    LL re=1;
    while(y)
    {
        if(y%2)
        {
            re=qiu(re,tem);
        }
        tem=qiu(tem,tem);
        y/=2;
    }
    return re;
}
int main()
{
    while(scanf("%I64d%I64d",&n,&p)!=EOF)
    {
        if(p==1)
            puts("0");
        else if(n==1)
        {
            puts("1");
        }
        else
        {
            printf("%I64d\n",((po(2,n)-2)%p+p)%p);
        }
    }
    return 0;
}