java.math.BigInteger.mod()方法的使用及代码示例

x33g5p2x  于2022-01-16 转载在 其他  
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本文整理了Java中java.math.BigInteger.mod()方法的一些代码示例，展示了BigInteger.mod()的具体用法。这些代码示例主要来源于Github/Stackoverflow/Maven等平台，是从一些精选项目中提取出来的代码，具有较强的参考意义，能在一定程度帮忙到你。BigInteger.mod()方法的具体详情如下：
包路径：java.math.BigInteger
类名称：BigInteger
方法名：mod

BigInteger.mod介绍

[英]Returns a BigInteger whose value is this mod m. The modulus m must be positive. The result is guaranteed to be in the interval [0, m) (0 inclusive, m exclusive). The behavior of this function is not equivalent to the behavior of the % operator defined for the built-in int's.
[中]返回一个BigInteger，其值为该模m。模m必须为正。结果保证在区间[0，m]（0包括，m不包括）内。此函数的行为与为内置int定义的%运算符的行为不同。

代码示例

代码示例来源：origin: stackoverflow.com

int nlen = n.bitLength();
BigInteger nm1 = n.subtract(BigInteger.ONE);
BigInteger r, s;
do {
  s = new BigInteger(nlen + 100, rnd);
  r = s.mod(n);
} while (s.subtract(r).add(nm1).bitLength() >= nlen + 100);
// result is in 'r'

代码示例来源：origin: stackoverflow.com

BigInteger val = new BigInteger("1254789363254125");
public boolean odd(BigInteger val) {
  if(!val.mod(new BigInteger("2")).equals(BigInteger.ZERO))
    return true;
  return false;
}

代码示例来源：origin: ethereum/ethereumj

@Override public Fp squared() { return new Fp(v.multiply(v).mod(P)); }
@Override public Fp dbl() { return new Fp(v.add(v).mod(P)); }

代码示例来源：origin: stackoverflow.com

public static void main(String... args) {
  BigInteger fact = fact(100);
  System.out.println("fact(100) = " + fact);
  System.out.println("fact(100).longValue() = " + fact.longValue());
  System.out.println("fact(100).intValue() = " + fact.intValue());
  int powerOfTwoCount = 0;
  BigInteger two = BigInteger.valueOf(2);
  while (fact.compareTo(BigInteger.ZERO) > 0 && fact.mod(two).equals(BigInteger.ZERO)) {
    powerOfTwoCount++;
    fact = fact.divide(two);
  }
  System.out.println("fact(100) powers of two = " + powerOfTwoCount);
}

private static BigInteger fact(long n) {
  BigInteger result = BigInteger.ONE;
  for (long i = 2; i <= n; i++)
    result = result.multiply(BigInteger.valueOf(i));
  return result;
}

代码示例来源：origin: i2p/i2p.i2p

protected byte[] engineSign() throws SignatureException {
  BigInteger elgp = key.getParams().getP();
  BigInteger pm1 = elgp.subtract(BigInteger.ONE);
  BigInteger elgg = key.getParams().getG();
  BigInteger x = ((ElGamalPrivateKey) key).getX();
  boolean ok;
  do {
    k = new BigInteger(2048, RandomSource.getInstance());
    ok = k.compareTo(pm1) == -1;
    ok = ok && k.compareTo(BigInteger.ONE) == 1;
    ok = ok && k.gcd(pm1).equals(BigInteger.ONE);
  } while (!ok);
  BigInteger r = elgg.modPow(k, elgp);
  BigInteger kinv = k.modInverse(pm1);
  BigInteger h = new NativeBigInteger(1, data);
  BigInteger s = (kinv.multiply(h.subtract(x.multiply(r)))).mod(pm1);

代码示例来源：origin: apache/zeppelin

BigInteger base = new BigInteger("" + DICTIONARY.length);
int exponent = 1;
BigInteger remaining = new BigInteger(value.toString());
while (true) {
 BigInteger b = remaining.mod(a); // 119 % 16 = 7 | 112 % 256 = 112
 BigInteger c = base.pow(exponent - 1);
 BigInteger d = b.divide(c);
 remaining = remaining.subtract(b); // 119 - 7 = 112 | 112 - 112 = 0
 if (remaining.equals(BigInteger.ZERO)) {
  break;

代码示例来源：origin: web3j/web3j

byte[] chainCode = Arrays.copyOfRange(i, 32, 64);
Arrays.fill(i, (byte) 0);
BigInteger ilInt = new BigInteger(1, il);
Arrays.fill(il, (byte) 0);
ECPoint ki = Sign.publicPointFromPrivate(ilInt).add(getPublicKeyPoint());
byte[] chainCode = Arrays.copyOfRange(i, 32, 64);
Arrays.fill(i, (byte) 0);
BigInteger ilInt = new BigInteger(1, il);
Arrays.fill(il, (byte) 0);
BigInteger privateKey = getPrivateKey().add(ilInt).mod(Sign.CURVE.getN());

代码示例来源：origin: DiUS/java-faker

private static String calculateIbanChecksum(String countryCode, String basicBankAccountNumber) {
  String basis = basicBankAccountNumber + countryCode + "00";
  StringBuilder sb = new StringBuilder();
  char[] characters = basis.toLowerCase().toCharArray();
  for (char c : characters) {
    if (Character.isLetter(c)) {
      sb.append(String.valueOf((c - 'a') + 10));
    } else {
      sb.append(c);
    }
  }
  int mod97 = new BigInteger(sb.toString()).mod(BigInteger.valueOf(97L)).intValue();
  return StringUtils.leftPad(String.valueOf(98 - mod97), 2, '0');
}

代码示例来源：origin: aws-amplify/aws-sdk-android

public AuthenticationHelper(String userPoolName) {
  do {
    a = new BigInteger(EPHEMERAL_KEY_LENGTH, SECURE_RANDOM).mod(N);
    A = GG.modPow(a, N);
  } while (A.mod(N).equals(BigInteger.ZERO));
  if (userPoolName.contains("_")) {
    poolName = userPoolName.split("_", 2)[1];
  } else {
    poolName = userPoolName;
  }
}

代码示例来源：origin: robovm/robovm

/**
 * Returns a {@code BigInteger} whose value is {@code
 * pow(this, exponent) mod modulus}. The modulus must be positive. The
 * result is guaranteed to be in the interval {@code [0, modulus)}.
 * If the exponent is negative, then
 * {@code pow(this.modInverse(modulus), -exponent) mod modulus} is computed.
 * The inverse of this only exists if {@code this} is relatively prime to the modulus,
 * otherwise an exception is thrown.
 *
 * @throws NullPointerException if {@code modulus == null} or {@code exponent == null}.
 * @throws ArithmeticException if {@code modulus < 0} or if {@code exponent < 0} and
 *     not relatively prime to {@code modulus}.
 */
public BigInteger modPow(BigInteger exponent, BigInteger modulus) {
  if (modulus.signum() <= 0) {
    throw new ArithmeticException("modulus.signum() <= 0");
  }
  int exponentSignum = exponent.signum();
  if (exponentSignum == 0) { // OpenSSL gets this case wrong; http://b/8574367.
    return ONE.mod(modulus);
  }
  BigInteger base = exponentSignum < 0 ? modInverse(modulus) : this;
  return new BigInteger(BigInt.modExp(base.getBigInt(), exponent.getBigInt(), modulus.getBigInt()));
}

代码示例来源：origin: i2p/i2p.i2p

@Test
public void squareAndDoubleReturnsCorrectResult() {
  for (int i=0; i<1000; i++) {
    // Arrange:
    final FieldElement f1 = getRandomFieldElement();
    final BigInteger b1 = toBigInteger(f1);
    // Act:
    final FieldElement f2 = f1.squareAndDouble();
    final BigInteger b2 = toBigInteger(f2).mod(getQ());
    // Assert:
    Assert.assertThat(b2, IsEqual.equalTo(b1.multiply(b1).multiply(new BigInteger("2")).mod(getQ())));
  }
}

代码示例来源：origin: i2p/i2p.i2p

public byte[] multiplyAndAdd(byte[] a, byte[] b, byte[] c) {
  return enc.encode(enc.toBigInteger(a).multiply(enc.toBigInteger(b)).add(enc.toBigInteger(c)).mod(l));
}

代码示例来源：origin: ethereum/ethereumj

@Override public Fp add(Fp o) { return new Fp(this.v.add(o.v).mod(P)); }
@Override public Fp mul(Fp o) { return new Fp(this.v.multiply(o.v).mod(P)); }

代码示例来源：origin: RUB-NDS/TLS-Attacker

public static BigInteger intceildiv(BigInteger c, BigInteger d) {
  if (c.mod(d).equals(BigInteger.ZERO)) {
    return intfloordiv(c, d);
  } else {
    return intfloordiv(c, d).add(BigInteger.ONE);
  }
}

代码示例来源：origin: i2p/i2p.i2p

BigInteger y1p = ELGPM1.subtract(a);
System.arraycopy(encrypted, ELG_HALF_LENGTH, buf, 0, ELG_HALF_LENGTH);
BigInteger d = new NativeBigInteger(1, buf);
BigInteger m = ya.multiply(d);
m = m.mod(CryptoConstants.elgp);
byte val[] = m.toByteArray();
int i;

代码示例来源：origin: ethereum/ethereumj

@Override public Fp sub(Fp o) { return new Fp(this.v.subtract(o.v).mod(P)); }
@Override public Fp squared() { return new Fp(v.multiply(v).mod(P)); }

代码示例来源：origin: java-json-tools/json-schema-validator

@Override
public void validate(final Processor<FullData, FullData> processor,
  final ProcessingReport report, final MessageBundle bundle,
  final FullData data)
  throws ProcessingException
{
  final BigInteger value
    = data.getInstance().getNode().bigIntegerValue();
  /*
   * We use a plain list to store failed divisors: remember that the
   * digested form was built with divisors in order, we therefore
   * only need insertion order, and a plain ArrayList guarantees that.
   */
  final List<BigInteger> failed = Lists.newArrayList();
  for (final BigInteger divisor: divisors)
    if (!value.mod(divisor).equals(BigInteger.ZERO))
      failed.add(divisor);
  if (failed.isEmpty())
    return;
  /*
   * There are missed divisors: report.
   */
  report.error(newMsg(data, bundle, "missingDivisors")
    .put("divisors", divisors).put("failed", failed));
}

代码示例来源：origin: jphp-group/jphp

BigInteger bigInt = new BigInteger(String.valueOf(value));
bigInt = bigInt.add(BIG_2_64);
while (bigInt.compareTo(BigInteger.ZERO) != 0) {
  int digit = bigInt.mod(BIG_TEN).intValue();

代码示例来源：origin: stackoverflow.com

BigInteger val = new BigInteger("1254789363254125");
public boolean odd(BigInteger val) {
  return !val.mod(new BigInteger("2")).equals(BigInteger.ZERO));
}

代码示例来源：origin: ethereum/ethereumj

void testAddMod(String v1, String v2, String v3) {
  DataWord dv1 = DataWord.of(Hex.decode(v1));
  DataWord dv2 = DataWord.of(Hex.decode(v2));
  DataWord dv3 = DataWord.of(Hex.decode(v3));
  BigInteger bv1 = new BigInteger(v1, 16);
  BigInteger bv2 = new BigInteger(v2, 16);
  BigInteger bv3 = new BigInteger(v3, 16);
  DataWord actual = dv1.addmod(dv2, dv3);
  BigInteger br = bv1.add(bv2).mod(bv3);
  assertEquals(actual.value(), br);
}

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