常用大数运算实例.doc

大数运算实例 BigInteger a=new BigInteger(“3”); BigInteger b=new BigInteger(“2”); BigInteger sum=a.add(b); //加 BigInteger difference=a.subtract(b); //减 BigInteger product=a.multiply(b); //乘 BigInteger quotient=a.divideb); //商 BigInteger rem=a.remainder(b); //余 1. 求商、余 ? public BigInteger[] divideAndRemainder(BigInteger val) throws ArithmeticException Since most division algorithms produce the quotient and the remainder at the same time, a more efficient way of capturing both of these values is provided by the divideAndRemainder() method. 实例. The answers are returned in an array of two BigIntegers, as follows: BigInteger a=new BigInteger(“9”); BigInteger b=new BigInteger(“2”); BigInteger[] answers=new BigInteger[2]; answers=a.divideAndRemainder(b); // , answers[0]=, answers[1] When this code completes, answers[0] contains the value 4 (as a BigInteger), and answers[1] contains 1. 2. 求幂 ? public BigInteger pow(int exponent) throws ArithmeticException This method returns a BigInteger whose value is thise where e = exponent and throws an ArithmeticException if e 0 (as the operation would yield a noninteger value). Note that e is an integer rather than a BigInteger 实例. Here is an example of how this method would be used (it calculates 2256): BigInteger base=new BigInteger(“2”); BigInteger humungous=base.pow(256); // 计算 3. 求最大公因子 ? public BigInteger gcd(BigInteger v) This method returns a BigInteger whose value is the greatest common divisor of |this| and |v|. It correctly returns (0, 0) as 0. 实例. BigInteger a, b,c; a = new BigInteger(18); b = new BigInteger(24); c =a.gcd(b); // 求 4. 模 ? public BigInteger mod(BigInteger m) This method returns a BigInteger whose value is this mod m. It throws an ArithmeticException if m ≤ 0. This method may return a negative value if the dividend is negative. 实例. BigInteger a, b, c; a = new BigInteger((100); b= new BigInteger(3); c = a.mod(b); //计算 5. 模幂 ? public BigInteger modPow(BigInteger e, BigInteger m) This method