zhx's contest(HDU-5187)(快速幂+快速乘法)

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

这题是推公式：

每一个位置其实可以是（a[1]............max..................min）

                                                或者是(max......................min...............a[n])

除了第一个位置和最后位置不符合其他都符合这种情况

其中一种是：2^(n-1)-2

两种是        ： 2^(n-1)-2

因为首尾多减了一次    所以要+2

ans=（2^(n-1)-2)×2+2;

        =（2^(n)-2+mod)%mod;

根据公式用快速幂加快速乘法：

#include<stdio.h>
#include<string.h>
typedef long long ll;
ll n,p;
ll qmul(ll a,ll b)
{
     ll ans=0;
     while(b)
     {
          if(b&1)
              ans=(ans+a)%p;
          a=(a*2)%p;
          b>>=1;
     }
     return ans;
}
ll qpow(ll a,ll b)
{
     ll ans=1;
     while(b)
     {
          if(b&1)
              ans=qmul(a,ans);
          a=qmul(a,a);
          b>>=1;
     }
     return ans;
}
int main()
{
     while(scanf("%lld%lld",&n,&p)!=EOF)
     {
         printf("%lld\n",(qpow(2,n)-2+p)%p);
     }
     return 0;
}