BestCoder Round #33(zhx's contest-快速乘法)

zhx's contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1366    Accepted Submission(s): 437

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 
    
    
    
    
     
   
   
   
   

 

Source

BestCoder Round #33

 

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本题使用快速乘法和快速幂

其中，快速乘法就是把乘数a的2进制(100010101)2 这样的数

分解成(2^p0+2^p1+...)*b=2^p0*b+2^p1*b ....  然后算

本题快速幂和快速乘法均用栈。
PS:由于忘了在sub()处%F 又WA又T的

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
typedef __int64 ll;
ll n,F;
ll add(ll a,ll b){return (a+b)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
ll mul(ll a,ll b) //c=a*b=a^l1+a^l2+...+a^l3 //把b改成2进制，各位分别相乘相加 
{
	ll c=0;
	while(b)
	{
		if (b&1) c=(c+a)%F; 
		a=(a+a)%F; 
		b>>=1; 
	}
	
	return c%F;
}
ll sub(ll a,ll b){if ((a-b)<0) return (a-b+F)%F;return (a-b)%F;}
ll pow2(ll a,ll b)
{
	if (b==1) return a%F;
	if (b==0) return 1%F;
	
	ll ret=1;
	while (b)
	{
		if (b&1) ret=mul(ret,a);
		a=mul(a,a);
		b>>=1;
	} 
	return ret%F;
	
}
int main()
{
//	freopen("zhx's contest.in","r",stdin);
	
	while(scanf("%I64d%I64d",&n,&F)==2)
	{
		if (n==1) printf("%I64d\n",1%F);
		else printf("%I64d\n",sub(pow2(2%F,n),2));
	}
	return 0;
}