HDU - 5187 - zhx's contest （快速幂+快速乘）

zhx's contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 448    Accepted Submission(s): 147

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

 

思路：由题意可以求出答案为(2^n-2)%p

但是n，p都是LL型的，快速幂的时候会爆LL，所以这里要用到快速乘法，快速乘法其实和快速幂差不多，就是把乘号改为加号

注意：当n为1时，要输出1，而当p为1时要输出0；

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long 
using namespace std;

LL n, p;

LL multi(LL a, LL b) {	//快速乘法，其实和快速幂差不多 
    LL ret = 0;
    while(b) {
        if(b & 1) ret = (ret + a) % p;
        a = (a + a) % p;
        b >>= 1;
    }
    return ret;
}

LL powmod(LL a, LL b) {	//快速幂 
    LL ret = 1;
    while(b) {
        if(b & 1) ret = multi(ret, a) % p;
        a = multi(a, a) % p;
        b >>= 1;
    }
    return ret;
}


int main() {
	while(cin >> n >> p) {
		if(p == 1) {
			cout << 0 << endl;
		} else if(n == 1) {
			cout << 1 << endl;
		} else {
			LL ans = powmod(2, n) - 2;
			if(ans < 0) ans += p;
			cout << ans << endl;
		}
	}
	return 0;
}