hdu 5187 基于快速幂的快速乘法

Posted 2020-11-19 vainglory

zhx‘s contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3568    Accepted Submission(s): 1146

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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快速乘法指的是在 a*b会爆ll 的情况下所进行的操作>>从而不必应用java大数来操作

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,p;
ll multi(ll a,ll b){
	ll res=0,tmp=a;
	while(b){
		if(b&1) (res+=tmp)%=p;
		b/=2;
		(tmp+=tmp)%=p;
	}
	return res;
}
ll pow_(ll a,ll b){
	ll res=1,tmp=a;
	while(b){
		if(b&1) res=multi(res,tmp);
		tmp=multi(tmp,tmp);
		b/=2;
	} 
	return res;
}
int main(){
	while(cin>>n>>p){
		ll ans=pow_(2,n);
		ans-=2;
		ans=(ans%p+p)%p;
		cout<<ans<<endl;
	}
	return 0;
}