Algorithm
Problem link- https://www.spoj.com/problems/GCD2/
GCD2 - GCD2
Frank explained its friend Felman the algorithm of Euclides to calculate the GCD of two numbers. Then Felman implements it algorithm
int gcd(int a, int b)
{
if (b==0)
return a;
else
return gcd(b,a%b);
}
and it proposes to Frank that makes it but with a little integer and another integer that has up to 250 digits.
Your task is to help Frank programming an efficient code for the challenge of Felman.
Input
The first line of the input file contains a number representing the number of lines to follow. Each line consists of two number A and B (0 <= A <= 40000 and A <= B < 10^250).
Output
Print for each pair (A,B) in the input one integer representing the GCD of A and B.
Example
Input: 2 2 6 10 11 Output: 2 1
Code Examples
#1 Code Example with Python Programming
Code -
Python Programming
def gcd(a, b):
while a > 0 and b > 0:
if a >= b:
a %= b
else:
b %= a
return a + b
T = int(input())
for i in range(T):
a, b = map(int, raw_input().split())
print(gcd(a, b))Input
2 6
10 11
Output
1
#2 Code Example with Java Programming
Code -
Java Programming
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Random;
public class GCD2 {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
Integer N = Integer.parseInt(br.readLine());
while (N-- > 0) {
String temp[] = br.readLine().split(" ");
Integer A = Integer.parseInt(temp[0]);
BigInteger B = new BigInteger(temp[1]);
if(A.equals(0)){
System.out.println("0");
continue;
}
Random randomGenerator = new Random();
while (true) {
Integer bb = B.mod(new BigInteger(A+"")).intValue();
long gcd1 = (gcd1(A, bb));
long big_made_small = specialMode(B.toString(), A);
long gcd2 = gcd2(A, big_made_small);
if(gcd1 != gcd2){
System.out.println(A+" " +bb);
System.out.println("gcd1 "+gcd1);
System.out.println("gcd2 "+gcd2);
break;
}
else{
System.out.println("PASS "+A+" "+B);
}
A = randomGenerator.nextInt(10000000);
B = new BigInteger(randomGenerator.nextInt(10000000)+"");
}
}
}
static int gcd1(int a, int b)
{
if (b==0)
return a;
else
return gcd1(b,a%b);
}
static long specialMode(String big,long small){
long remainder = 0;
for (int index = 0; index < big.length(); index++){
if(remainder >= small){
remainder = remainder % small;
}
remainder = remainder * 10;
remainder += Integer.parseInt(big.charAt(index)+ "");
}
return remainder % small;
}
static long gcd2(long A,long B){
if(B == 0 ){
return A;
}
return gcd2(B,A % B);
}
}Input
2 6
10 11
Output
1
#3 Code Example with C++ Programming
Code -
C++ Programming
#include<stdio.h>
#include<string.h>
char b[252];
int gcd(int a, int b){
if(!a) return b;
return gcd(b%a,a);
}
int main(){
int t,a,mod,i,l;
for(scanf("%d",&t);t--;){
mod = 0;
scanf("%d",&a);
scanf("%s",b);
if(!a){
printf("%s\n",b);
continue;
}
l = strlen(b);
for(i=0;i<l;i++){
mod = (mod*10 + b[i] - '0')%a;
}
printf("%d\n",gcd(mod,a));
}
return 0;
}Input
2 6
10 11
Output
1
Demonstration
SPOJ Solution-GCD2-Solution in C, C++, Java, Python