RSA算法原理

这两篇博文讲解的真是细致：

http://www.ruanyifeng.com/blog/2013/06/rsa_algorithm_part_one.html

http://www.ruanyifeng.com/blog/2013/07/rsa_algorithm_part_two.html?20151007165925

上面是原文出处，可能会有一些图片显示不出来，这里再贴一个转发地址：

http://blog.jobbole.com/42699/

 

这里贴一个应用：

ural  RSA Attack

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

 

Description

The RSA problem is the following: given a positive integer n that is a product of two distinct odd primes p and q, a positive integer e such that gcd(e, (p-1)*(q-1)) = 1, and an integer c, find an integer m such that m ^e = c (mod n).

Input

One number K (K <= 2000) in the first line is an amount of tests. Each next line represents separate test, which contains three positive integer numbers – e, n and c (e, n, c <= 32000, n = p*q, p, q – distinct odd primes, gcd(e, (p-1)*(q-1)) = 1, e < (p-1)*(q-1) ).

Output

For each input test the program must find the encrypted integer m.

Sample Input

input	output
3 9 187 129 11 221 56 7 391 204	7 23 17

AC代码：

#include <iostream>
#include <cmath>
using namespace std;


//a^b%c 
int fastmod(int a,int b,int c){
    int num=1;
    a%=c;//这里不进行初始化也是可以的
    
    while(b>0){
        if(b%2==1){
            num=(num*a)%c;    
        }
        b/=2;     //这一步将b->log2(b) 
        a=(a*a)%c;
    }
    return num;
}

int getp(int n){   //得到大于1的最小的因数 
    int p=2;
    while(n%p!=0){
        p++;
    }
    return p;
}
int e_gcd(int a,int b,int &x,int &y){
    if(b==0){
        x=1;
        y=0;
        return a;
    }
    int ans=e_gcd(b,a%b,x,y);
    int temp=x;
    x=y;
    y=temp-a/b*y;
    return ans;
} 

int cal(int a,int m){
    int x,y;
    int gcd=e_gcd(a,m,x,y);
    if(1%gcd!=0)  return -1;
    x*=1/gcd;
    m=abs(m);
    int ans=x%m;
    if(ans<=0) ans+=m;
    return ans;
}


int main(){   
    int k;
    cin>>k;
    while(k--){
        int e,n,c;
        cin>>e>>n>>c; 
        int p=getp(n),q=n/p;
    
        int d=cal(e,(p-1)*(q-1));    
        
        cout<<fastmod(c,d,n)<<endl;    
    }
}