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

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本文探讨了一个关于问题难度排列的问题，zhx需要为他的后辈准备n个不同难度的问题，并希望这些问题按难度排列时满足特定的美学标准。文章详细介绍了如何通过快速乘法和快速幂的方法来解决这一问题，最终求出合法排列的数量。

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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hujie | We have carefully selected several similar problems for you: 5193 5192 5191 5190 5189

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;
}