POJ 2447 RSA —— RSA加密算法的破解过程

RSA

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 3358 Accepted: 709

Description

RSA is the best-known public key encryption algorithm. In this algorithm each participant has a private key that is shared with no one else and a public key which is published so everyone knows it. To send a secure message to this participant, you encrypt the message using the widely known public key; the participant then decrypts the messages using his or her private key. Here is the procedure of RSA: 

First, choose two different large prime numbers P and Q, and multiply them to get N (= P * Q). 
Second, select a positive integer E (0 < E < N) as the encryption key such that E and T= (P - 1) * (Q - 1) are relatively prime. 
Third, compute the decryption key D such that 0 <= D < T and (E * D) mod T = 1. Here D is a multiplicative inverse of E, modulo T. 

Now the public key is constructed by the pair {E, N}, and the private key is {D, N}. P and Q can be discarded. 

Encryption is defined by C = (M ^ E) mod N, and decryption is defined by M = (C ^ D) mod N, here M, which is a non-negative integer and smaller than N, is the plaintext message and C is the resulting ciphertext. 

To illustrate this idea, let’s see the following example: 
We choose P = 37, Q = 23, So N = P * Q = 851, and T = 792. If we choose E = 5, D will be 317 ((5 * 317) mod 792 = 1). So the public key is {5, 851}, and the private key is {317, 851}. For a given plaintext M = 7, we can get the ciphertext C = (7 ^ 5) mod 851 = 638. 

As we have known,for properly choosen very large P and Q, it will take thousands of years to break a key, but for small ones, it is another matter. 

Now you are given the ciphertext C and public key {E, N}, can you find the plaintext M?

Input

The input will contain several test cases. Each test case contains three positive integers C, E, N (0 < C < N, 0 < E < N, 0 < N < 2 ^ 62).

Output

Output the plaintext M in a single line.

Sample Input

638 5 851

Sample Output

7

Source

POJ Monthly,static

一道数论的模板题，做法很简单，但是需要一个比较复杂的证明。

关于RSA请参考http://en.wikipedia.org/wiki/RSA_(algorithm)

算法举例：

我们只要模拟这一过程即可。

#include <cstdio>#include <cmath>#include <algorithm>#include <iostream>#include <cstring>#include <map>#include <string>#include <stack>#include <cctype>#include <vector>#include <queue>#include <set>#include <utility>#include <cassert>using namespace std;///#define Online_Judge#define outstars cout << "***********************" << endl;#define clr(a,b) memset(a,b,sizeof(a))#define lson l , mid  , rt << 1#define rson mid + 1 , r , rt << 1 | 1#define mk make_pair#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)const int MAXN = 40000 + 50;const int MAXS = 10000 + 50;const int sigma_size = 26;const long long LLMAX = 0x7fffffffffffffffLL;const long long LLMIN = 0x8000000000000000LL;const int INF = 0x7fffffff;const int IMIN = 0x80000000;const int inf = 1 << 30;#define eps 1e-10const long long MOD = 1000000000 + 7;const int mod = 10007;typedef long long LL;const double PI = acos(-1.0);typedef double D;typedef pair<int , int> pii;typedef vector<int> vec;typedef vector<vec> mat;#define Bug(s) cout << "s = " << s << endl;///#pragma comment(linker, "/STACK:102400000,102400000")#define gcc 10007inline __int64 gcd(__int64 a , __int64 b){    if(b > a)return gcd(b , a);    return b == 0 ? a : gcd(b , a % b);}inline __int64 Produce_Mod(__int64 a , __int64 b , __int64 Mod){    __int64 sum = 0;    while(b > 0)    {        if(b & 1)sum = (sum + a) % Mod;        a = (a + a) % Mod;        b >>= 1;    }    return sum;}inline __int64 Power(__int64 a , __int64 b , __int64 Mod){    __int64 sum = 1;    while(b > 0)    {        if(b & 1)sum = Produce_Mod(sum , a , Mod);        a = Produce_Mod(a , a , Mod);        b >>= 1;    }    return sum;}__int64 Pollard_rho(__int64 n){    int i = 1;    __int64 x = rand() % (n - 1) + 1;    __int64 y = x;    __int64 k = 2;    __int64 d;    do    {        i++;        d = gcd(n + y - x , n);        if(d > 1 && d < n)        {            return d;        }        if(i == k)y = x , k *= 2;        x = ((Produce_Mod(x , x , n) - gcc) % n + n) % n;    }while(y != x);    return n;}__int64 extgcd(__int64 a , __int64 b , LL &x , LL &y){    if(b == 0){x = 1;y = 0; return a;}    __int64 d = extgcd(b , a % b , x , y);    LL t = x ; x = y ; y = t - a / b * y;    return d;}int main(){    //ios::sync_with_stdio(false);    #ifdef Online_Judge        freopen("in.txt","r",stdin);        freopen("out.txt","w",stdout);    #endif // Online_Judge    __int64 c , e , n , p , q ,m , t , d;    while(~scanf("%lld%lld%lld" , &c , &e , &n))    {        p = Pollard_rho(n);q = n / p;        t = (p - 1) * (q - 1);        LL x;        extgcd(e , t , d , x);        d = (d % t + t) % t;        printf("%lld\n" , Power(c , d , n));    }    return 0;}