Problem 3- The Power of Cryptography

Power of Cryptography

Background

Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers modulo functions of these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered to be of only theoretical interest. This problem involves the efficient computation of integer roots of numbers.

The Problem

Given an integer and an integer you are to write a program that determines , the positive root of p. In this problem, given such integers n and p, p will always be of the form for an integer k (this integer is what your program must find).

The Input

The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs , and there exists an integer k, such that .

The Output

For each integer pair n and p the value should be printed, i.e., the number k such that .

Sample Input

2
16
3
27
7
4357186184021382204544

Sample Output

4
3
1234

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scroll down for solution working till 10^18 but not more than that






























































//--------Main class

package

crypt;

import

java.io.InputStreamReader;

import

java.util.ArrayList;

import

java.util.Scanner;

public

class Main {

/**

*

@param args

*/

public static void main(String[] args) {

long n, p;

int k;

String more =

"n";

Scanner scan =

new Scanner(new InputStreamReader(System.in));

ArrayList<Long> nlist =

new ArrayList<Long>();

ArrayList<Long> plist =

new ArrayList<Long>();

ArrayList<Integer> klist =

new ArrayList<Integer>();

System.

out.println("Enter Numbers Less than 10^18");

do {

System.

out.println("Enter n ,p:");

n = scan.nextLong();

p = scan.nextLong();

nlist.add(n);

plist.add(p);

System.

out.println("MORE ?(y)");

more = scan.next();

}

while (more.equalsIgnoreCase("y"));

System.

out.println("USER INPUT");

for (int i = 0; i < nlist.size(); i++) {

System.

out.println(nlist.get(i) + " " + plist.get(i));

k = Divider.getPower(n, p);

klist.add(k);

}

System.

out.println("OUTPUT POWERS");

for (Integer integer : klist) {

System.

out.println(integer);

}

}

}

//------------------2nd class

package

crypt;

public

class Divider {

public static int getPower(long n, long p) {

int k = 0;

while (p > 1) {

p /= n;

k++;

}

return k;

}

}

//-----------EOP