CODEWITHSUNDEEP
java
Chid
Chid

Reputation: 19

Java modulo operation using for loop

How do I carry out the following modulo operation in java.

5^7 mod 11 = 5^(2+2+2+1) mod 11.

I tried using for loop, but I am not able to get the required output

Upvotes: 0

Views: 1664

Answers (3)

Joop Eggen
Joop Eggen

Reputation: 109567

As this seems homework, I'll give the recursive version of what might be intended: using the exponent's binary decomposition:

x^(2n) = q * q where q = 2^n

int pow(int base, int exp, int modulo) {
    if (exp == 0) {
        return 1;
    }
    base %= modulo;
    if (exp == 1) {
        return base;
    }
    if (exp % 2 == 1) {
        return (base * pow(base, exp, modulo)) % modulo;
    }
    int sqroot = pow(base, exp / 2, modulo) % modulo;
    return (sqroot * sqroot) % modulo;
}

int result = pow(5, 7, 11);

As the modulo is a prime, as is the base, there are several optimizations possible.

Upvotes: 0

OldCurmudgeon
OldCurmudgeon

Reputation: 65821

Here's both ways:

public void testMath(int x, int y, int z) {
    double pow = Math.pow(x, y);
    double mod = pow % z;
    System.out.println("Math - pow(" + x + "," + y + "," + z + ") = " + pow + " mod = " + mod);
}

public void testLoop(int x, int y, int z) {
    double v = 1;
    for (int i = 0; i < y; i++) {
        v *= x;
        v %= z;
    }
    System.out.println("Math - pow(" + x + "," + y + "," + z + ") = " + " mod = " + v);
}

public void test() {
    testMath(5, 7, 11);
    testLoop(5, 7, 11);
}

Upvotes: 0

Ren&#233; Winkler
Ren&#233; Winkler

Reputation: 7068

You don't need to use a loop for this operation. You can just use:

Math.pow(5.0,7.0) % 11.0

Upvotes: 1

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