蒙哥马利幂模运算

蒙哥马利模乘的优点在于减少了取模的次数（在大数的条件下）以及简化了除法的复杂度（在2的k次幂的进制下除法仅需要进行左移操作）。模幂运算是RSA 的核心算法，最直接地决定了RSA 算法的性能。

针对快速模幂运算这一课题，西方现代数学家提出了大量的解决方案，通常都是先将幂模运算转化为乘模运算。

例如求D=C^15%N

由于：a*b % n = (a % n)*(b % n) % n

所以令：

C1 =C*C % N =C^2 % N

C2 =C1*C % N =C^3 % N

C3 =C2*C2 % N =C^6 % N

C4 =C3*C % N =C^7 % N

C5 =C4*C4 % N =C^14 % N

C6 =C5*C % N =C^15 % N

int Monto(int a,int b,int c){int ans = 1;while(b){if(b&1)           ans = (ans*a)%c;b>>=1;a =(a*a)%c;}return ans;}

例如HDU 1395

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15203    Accepted Submission(s): 4699

Problem Description

Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.

 

Input

One positive integer on each line, the value of n.

 

Output

If the minimum x exists, print a line with 2^x mod n = 1.

Print 2^? mod n = 1 otherwise.

You should replace x and n with specific numbers.

 

Sample Input

25

 

Sample Output

2^? mod 2 = 12^4 mod 5 = 1

 

Author

MA, Xiao

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;int Monto(int a,int b,int c){int ans = 1;while(b){if(b&1)           ans = (ans*a)%c;b>>=1;a =(a*a)%c;}return ans;}int main(){int i,j,m,n,g,t,d,k;int cnt,a,b,c;while(cin>>n){if(n%2==0 || n==1){cout<<"2^? mod "<<n<<" = 1"<<endl;}else{for(i=1;;i++)if(Monto(2,i,n)==1){cout<<"2^"<<i<<" mod "<<n<<" = 1"<<endl;break;}}}}

虽然这样可以AC，但是，挺耗时的，还不如暴力的循环