ural 1141. RSA Attack RSA算法

1141. RSA Attack

Time limit: 1.0 second
Memory limit: 64 MB

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

Problem Author: Michael Medvedev

思路：这个主要是一个关于RSA加密算法的问题，如果想具体了解可以看阮一峰的网络日志

该题的解法很简单， 首先把n分解成两个质数p, q;

然后求e与（p-1）*（q-1）的逆元d

最后求一发 c^d (mod n)

#include <iostream>
#include <cstdio>
using namespace std;
const int MAXN = 32000;
int N, T;
int prime[MAXN+10], cnt = 0;
bool a[MAXN+10];
int extgcd(int a, int b, int &x, int &y){
    int d = a;
    if(b != 0){
        d = extgcd(b, a%b, y, x);
        y -= (a/b)*x;
    }else {
        x = 1, y = 0;
    }return d;
}

int mod_inv(int a, int m){
    int x, y;
    extgcd(a, m, x, y);
    return (m+x%m)%m;
}

void init(){
    for(int i = 2; i <= MAXN; i++) a[i] = true;
    for(int i = 2; i <= MAXN; i++){
        if(a[i]){
            prime[++cnt] = i;
        }
        for(int j = 1; j <= MAXN; j++){
            if(prime[j]*i > MAXN) break;
            a[prime[j]*i] = false;
            if(i%prime[j] == 0) break;
        }
    }
}
int quick_pow(int a, int x, int p){
    int ans = 1;
    while(x > 0){
        if(x&1){
            x--;
            ans = ans*a%p;
        }
        x >>= 1;
        a = a*a%p;
    }
    return ans;
}

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    init();
    scanf("%d", &T);
    while(T--){
        int e, p, q, n, c, m, d;
        scanf("%d%d%d", &e, &n, &c);
        for(int i = 1; i <= cnt && prime[i] <= n; i++){
            if(prime[i]%2 == 1 && n % prime[i] == 0){
                p = prime[i];
                q = n/prime[i];
                break;
            }
        }
        d = mod_inv(e, (p-1)*(q-1));
        m = quick_pow(c, d, n);
        printf("%d
", m);
    }
    return 0;
}