我正在尝试实施Modular Exponentiation,但我无法得到正确答案:
public static BigInteger modPow(BigInteger b,BigInteger e,BigInteger m)
{//计算模幂指数并返回BigInteger类的对象
BigInteger x= new BigInteger("1"); //The default value of x
BigInteger power ;
power=b.mod(m);
String t =e.toString(2); //convert the power to string of binary
String reverse = new StringBuffer(t).reverse().toString();
for (int i=0;i<reverse.length();i++ ) { //this loop to go over the string char by char by reverse
if(reverse.charAt(i)=='1') { //the start of if statement when the char is 1
x=x.multiply(power);
x=x.mod(m);
power=power.multiply(power);
power=power.mod(m);
} //the end of if statement
}//the end of for loop
return x;
} //the end of the method modPow
答案 0 :(得分:1)
对于零指数位,您不做任何事情。如果指数为2 0 且指数为2 2048 ,你会得到相同的结果吗?
这些语句应该来自if
子句,并且在循环的每次迭代中执行,无论该位是0还是1:
power=power.multiply(power);
power=power.mod(m);
此外,使用e.testBit(i)
迭代指数的位将更有效,更容易理解。即使不允许使用modPow()
,testBit()
也应该没问题。
这是我的版本,包括修复bug和我建议摆脱字符串转换。对于一般数字,它似乎也可靠地工作。它不处理负指数和其他一些特殊情况。
public class CrazyModPow
{
public static void main(String[] argv)
{
for (int count = 1; true; ++count) {
Random rnd = new Random();
BigInteger base = BigInteger.probablePrime(512, rnd);
BigInteger exp = BigInteger.probablePrime(512, rnd);
BigInteger mod = BigInteger.probablePrime(1024, rnd);
if (!base.modPow(exp, mod).equals(modPow(base, exp, mod))) {
System.out.println("base: " + base);
System.out.println("exp: " + exp);
System.out.println("mod: " + mod);
}
else if ((count % 10) == 0) {
System.out.printf("Tested %d times.%n", count);
}
}
}
public static BigInteger modPow(BigInteger base, BigInteger e, BigInteger m)
{
BigInteger result = BigInteger.ONE;
base = base.mod(m);
for (int idx = 0; idx < e.bitLength(); ++idx) {
if (e.testBit(idx)) {
result = result.multiply(base).mod(m);
}
base = base.multiply(base).mod(m);
}
return result;
}
}