RSA解密

题目

Alice decides to use RSA with the public key N = 1889570071. In order to guard against transmission errors, Alice has Bob encrypt his message twice, once using the encryption exponent e1 = 1021763679 and once using the encryption exponent e2 = 519424709. Eve intercepts the two encrypted messages c1 = 1244183534 and c2 = 732959706. Assuming that Eve also knows N and the two encryption exponents e1 and e2. Please help Eve recover Bob’s plaintext without finding a factorization of N.

解题过程

由题可知：Eve知道N、C1、C2、e1、e2，要破解M

∵ C1≡ Me1mod N

C2≡ Me2mod N

∵ gcd(e1,e2) = 1//此处说明两者互质

∴ ∃ x , y ,//根基欧几里得扩充定理可知

e1* x + e2* y = 1

∴ C1x* C2y≡ Me1x+e2ymod N

∴ C1x* C2y≡ M mod N

∵ x > 0 , y < 0

∴ C1x* C2y= C1x* (C2-1)-y

C2-1即为 C2在模N乘法的逆元

∴ M = [(C1x) mod n * (C2-1)-ymod n] mod n

∴ M = 1054592380

此为解题思路

代码如下：

#include

using namespace std;

typedef long long L;

L exgcd(L a, L b, L &x, L &y)//ax+by=1返回a,b的gcd，同时求的一组满足题目的最小正整数解，也就是解题思路中的x,y

{

L ans, t;

if (b == 0)

{

x = 1;

y = 0;

return a;

}

ans = exgcd(b, a%b, x, y);

t = x;

x = y;

y = t - (a / b)*y;

return ans;

}

L expmod(L a, L b, L n)//快速幂求余算法a^b%n

{

L res = 1;

a %= n;

while (b)

{

if (b & 1)//b是否为奇数

res = (res * a) % n;

a = (a * a) % n;

b /= 2;

}

return res;

}

void main()

{

L n = 1889570071;

L e1 = 1021763679;

L e2 = 519424709;

L c1 = 1244183534;

L c2 = 732959706;

L m, u, v, t, t1, t2, t3;

exgcd(e1, e2, u, v);

cout << u << endl << v << endl;

t1 = expmod(c1, u, n);

t2 = expmod(c2, v, n);

t3 = t1 * t2;

t = expmod(t3, 1, n);

m = t;

cout << m << endl;

}