public RSA(Random random) {
savedRandom = random;
p = getPrime(random); // p and q are arbitrary large primes
q = getPrime(random);
n = p.multiply(q);
phiN = (p.subtract(ONE)).multiply(q.subtract(ONE));
S = getPrime(random); // s is an arbitrary secret; we'll use a prime
V = S.modPow(new BigInteger("2"), n);
}
private void makeKeys() {
PrimeGenerator pg = new PrimeGenerator(513, 10, sr);
do {
p = pg.getStrongPrime();
} while (p.subtract(BigIntegerMath.ONE).mod(BigIntegerMath.THREE).equals(BigIntegerMath.ZERO));
do {
q = pg.getStrongPrime();
} while (q.subtract(BigIntegerMath.ONE).mod(BigIntegerMath.THREE).equals(BigIntegerMath.ZERO));
modulus = p.multiply(q);
// Use 3 as enciphering exponent - OK since we are using salt
decipherExp =
BigIntegerMath.THREE.modInverse(
p.subtract(BigIntegerMath.ONE).multiply(q.subtract(BigIntegerMath.ONE)));
ciphertextBlockSize = (modulus.bitLength() - 1) / 8 + 1;
// Maximum size of plaintext is 6 bytes less than ciphertext
// 1 to get under modulus
// 4 for the salt
// 1 for a pad byte
plaintextBlockSize = ciphertextBlockSize - 6;
}
void start() throws IOException {
BigInteger n = new BigInteger(reader.next());
String s = reader.next();
BigInteger answer = BigInteger.ONE;
while (n.compareTo(BigInteger.ONE) > 0) {
answer = answer.multiply(n);
n = n.subtract(BigInteger.valueOf(s.length()));
}
writer.println(answer);
}
void solve(int caseNum) {
int n = in.nextInt();
BigInteger[] past = new BigInteger[n], d = new BigInteger[n];
for (int i = 0; i < n; i++) past[i] = new BigInteger(in.next());
Arrays.sort(past);
for (int i = 0; i < n; i++) d[i] = past[i].subtract(past[0]);
for (int i = 2; i < n; i++) d[i] = d[i].gcd(d[i - 1]);
BigInteger gcd = d[n - 1], mod = past[0].mod(gcd);
System.out.printf("Case #%d: ", caseNum);
if (mod.equals(BigInteger.ZERO)) System.out.println(0);
else System.out.println(gcd.subtract(mod));
}
