codeparrot/xlcost-text-to-code huggingface dataset:
Divide the two given numbers by their common divisors | Java implementation of above approach ; Function to calculate gcd of two numbers ; Function to calculate all common divisors of two given numbers a , b -- > input integer numbers ; find gcd of a , b ; Driver code
class GFG { static int gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } static void commDiv ( int a , int b ) { int n = gcd ( a , b ) ; a = a / n ; b = b / n ; System . out . println ( " A ▁ = ▁ " + a + " , ▁ B ▁ = ▁ " + b ) ; } public static void main ( String [ ] args ) { int a = 10 , b = 15 ; commDiv ( a , b ) ; } }
Minimum absolute difference between N and a power of 2 | Java implementation of the above approach ; Function to return the minimum difference between N and a power of 2 ; Power of 2 closest to n on its left ; Power of 2 closest to n on its right ; Return the minimum abs difference ; Driver code
class GFG { static int minAbsDiff ( int n ) { int left = 1 << ( ( int ) Math . floor ( Math . log ( n ) / Math . log ( 2 ) ) ) ; int right = left * 2 ; return Math . min ( ( n - left ) , ( right - n ) ) ; } public static void main ( String [ ] args ) { int n = 15 ; System . out . println ( minAbsDiff ( n ) ) ; } }
Find probability that a player wins when probabilities of hitting the target are given | Java implementation of the approach ; Function to return the probability of the winner ; Driver Code ; Will print 9 digits after the decimal point
import java . util . * ; import java . text . DecimalFormat ; class solution { static double find_probability ( double p , double q , double r , double s ) { double t = ( 1 - p / q ) * ( 1 - r / s ) ; double ans = ( p / q ) / ( 1 - t ) ; return ans ; } public static void main ( String args [ ] ) { double p = 1 , q = 2 , r = 1 , s = 2 ; DecimalFormat dec = new DecimalFormat ( " # 0.000000000" ) ; System . out . println ( dec . format ( find_probability ( p , q , r , s ) ) ) ; } }
Represent n as the sum of exactly k powers of two | Set 2 | Java implementation of the above approach ; Function to print k numbers which are powers of two and whose sum is equal to n ; Initialising the sum with k ; Initialising an array A with k elements and filling all elements with 1 ; Iterating A [ ] from k - 1 to 0 ; Update sum and A [ i ] till sum + A [ i ] is less than equal to n ; Impossible to find the combination ; Possible solution is stored in A [ ] ; Driver code
import java . util . Arrays ; public class GfG { public static void FindAllElements ( int n , int k ) { int sum = k ; int [ ] A = new int [ k ] ; Arrays . fill ( A , 0 , k , 1 ) ; for ( int i = k - 1 ; i >= 0 ; -- i ) { while ( sum + A [ i ] <= n ) { sum += A [ i ] ; A [ i ] *= 2 ; } } if ( sum != n ) { System . out . print ( " Impossible " ) ; } else { for ( int i = 0 ; i < k ; ++ i ) System . out . print ( A [ i ] + " ▁ " ) ; } } public static void main ( String [ ] args ) { int n = 12 ; int k = 6 ; FindAllElements ( n , k ) ; } }
Check whether a + b = c or not after removing all zeroes from a , b and c | Java program to check the sum after Removing all zeroes is true or not ; Function to remove zeroes ; Initialize result to zero holds the Result after removing zeroes from no ; Initialize variable d to 1 that holds digits of no ; Loop while n is greater then zero ; Check if n mod 10 is not equal to zero ; store the result by removing zeroes And increment d by 10 ; Go to the next digit ; Return the result ; Function to check if sum is true after Removing all zeroes . ; Call removeZero ( ) for both sides and check whether they are equal After removing zeroes . ; Driver code
public class GfG { public static int removeZero ( int n ) { int res = 0 ; int d = 1 ; while ( n > 0 ) { if ( n % 10 != 0 ) { res += ( n % 10 ) * d ; d *= 10 ; } n /= 10 ; } return res ; } public static boolean isEqual ( int a , int b ) { if ( removeZero ( a ) + removeZero ( b ) == removeZero ( a + b ) ) return true ; return false ; } public static void main ( String [ ] args ) { int a = 105 , b = 106 ; if ( isEqual ( a , b ) == true ) System . out . println ( " Yes " ) ; else System . out . println ( " No " ) ; } }
A Sum Array Puzzle | Java implementation of above approach ; Allocate memory for temporary arrays leftSum [ ] , rightSum [ ] and Sum [ ] ; Left most element of left array is always 0 ; Right most element of right array is always 0 ; Construct the left array ; Construct the right array ; Construct the sum array using left [ ] and right [ ] ; print the sum array ; Driver function to test above function
import java . util . * ; import java . lang . * ; import java . io . * ; class Geeks { public static void sumArray ( int arr [ ] , int n ) { int leftSum [ ] = new int [ n ] ; int rightSum [ ] = new int [ n ] ; int Sum [ ] = new int [ n ] ; int i = 0 , j = 0 ; leftSum [ 0 ] = 0 ; rightSum [ n - 1 ] = 0 ; for ( i = 1 ; i < n ; i ++ ) leftSum [ i ] = arr [ i - 1 ] + leftSum [ i - 1 ] ; for ( j = n - 2 ; j >= 0 ; j -- ) rightSum [ j ] = arr [ j + 1 ] + rightSum [ j + 1 ] ; for ( i = 0 ; i < n ; i ++ ) Sum [ i ] = leftSum [ i ] + rightSum [ i ] ; for ( i = 0 ; i < n ; i ++ ) System . out . print ( Sum [ i ] + " ▁ " ) ; } public static void main ( String [ ] args ) { int arr [ ] = { 3 , 6 , 4 , 8 , 9 } ; int n = arr . length ; sumArray ( arr , n ) ; } }
Find minimum x such that ( x % k ) * ( x / k ) == n | Set | Java Program to find the minimum positive X such that the given equation holds true ; This function gives the required answer ; Iterate for all the factors ; Check if i is a factor ; Consider i to be A and n / i to be B ; Consider i to be B and n / i to be A ; Driver Code to test above function
import java . util . * ; class solution { static int minimumX ( int n , int k ) { int mini = Integer . MAX_VALUE ; for ( int i = 1 ; i * i <= n ; i ++ ) { if ( n % i == 0 ) { int fir = i ; int sec = n / i ; int num1 = fir * k + sec ; int res = ( num1 / k ) * ( num1 % k ) ; if ( res == n ) mini = Math . min ( num1 , mini ) ; int num2 = sec * k + fir ; res = ( num2 / k ) * ( num2 % k ) ; if ( res == n ) mini = Math . min ( num2 , mini ) ; } } return mini ; } public static void main ( String args [ ] ) { int n = 4 , k = 6 ; System . out . println ( minimumX ( n , k ) ) ; n = 5 ; k = 5 ; System . out . println ( minimumX ( n , k ) ) ; } }
Find minimum x such that ( x % k ) * ( x / k ) == n | Java Program to find the minimum positive X such that the given equation holds true ; This function gives the required answer ; Iterate over all possible remainders ; it must divide n ; Driver Code to test above function
class Solution { static int minimumX ( int n , int k ) { int ans = Integer . MAX_VALUE ; for ( int rem = k - 1 ; rem > 0 ; rem -- ) { if ( n % rem == 0 ) ans = Math . min ( ans , rem + ( n / rem ) * k ) ; } return ans ; } public static void main ( String args [ ] ) { int n = 4 , k = 6 ; System . out . println ( minimumX ( n , k ) ) ; n = 5 ; k = 5 ; System . out . println ( minimumX ( n , k ) ) ; } }
Find nth Hermite number | Java program to find nth Hermite number ; Function to return nth Hermite number ; Base condition ; Driver Code ; Print nth Hermite number
import java . util . * ; class GFG { static int getHermiteNumber ( int n ) { if ( n == 0 ) return 1 ; else if ( n == 1 ) return 1 ; else return - 2 * ( n - 1 ) * getHermiteNumber ( n - 2 ) ; } public static void main ( String [ ] args ) { int n = 6 ; System . out . println ( getHermiteNumber ( n ) ) ; } }
Find numbers a and b that satisfy the given conditions | Java implementation of the above approach ; Function to print the required numbers ; Suppose b = n and we want a % b = 0 and also ( a / b ) < n so a = b * ( n - 1 ) ; Special case if n = 1 we get a = 0 so ( a * b ) < n ; If no pair satisfies the conditions ; Driver code
public class GFG { static void find ( int n ) { int b = n ; int a = b * ( n - 1 ) ; if ( a * b > n && a / b < n ) { System . out . print ( " a ▁ = ▁ " + a + " , ▁ b ▁ = ▁ " + b ) ; } else System . out . println ( - 1 ) ; } public static void main ( String [ ] args ) { int n = 10 ; find ( n ) ; } }
Closest perfect square and its distance | Java program to find the closest perfect square taking minimum steps to reach from a number ; Function to check if a number is perfect square or not ; Function to find the closest perfect square taking minimum steps to reach from a number ; Variables to store first perfect square number above and below N ; Finding first perfect square number greater than N ; Finding first perfect square number less than N ; Variables to store the differences ; Driver code
class GFG { static boolean isPerfect ( int N ) { if ( ( Math . sqrt ( N ) - Math . floor ( Math . sqrt ( N ) ) ) != 0 ) return false ; return true ; } static void getClosestPerfectSquare ( int N ) { if ( isPerfect ( N ) ) { System . out . println ( N + " ▁ " + "0" ) ; return ; } int aboveN = - 1 , belowN = - 1 ; int n1 ; n1 = N + 1 ; while ( true ) { if ( isPerfect ( n1 ) ) { aboveN = n1 ; break ; } else n1 ++ ; } n1 = N - 1 ; while ( true ) { if ( isPerfect ( n1 ) ) { belowN = n1 ; break ; } else n1 -- ; } int diff1 = aboveN - N ; int diff2 = N - belowN ; if ( diff1 > diff2 ) System . out . println ( belowN + " ▁ " + diff2 ) ; else System . out . println ( aboveN + " ▁ " + diff1 ) ; } public static void main ( String args [ ] ) { int N = 1500 ; getClosestPerfectSquare ( N ) ; } }
Fraction | Java program to add 2 fractions ; Function to return gcd of a and b ; Function to convert the obtained fraction into it 's simplest form ; Finding gcd of both terms ; Converting both terms into simpler terms by dividing them by common factor ; Function to add two fractions ; Finding gcd of den1 and den2 ; Denominator of final fraction obtained finding LCM of den1 and den2 LCM * GCD = a * b ; Changing the fractions to have same denominator . Numerator of the final fraction obtained ; Calling function to convert final fraction into it 's simplest form ; Driver Code
import java . util . * ; class GFG { static int den3 , num3 ; static int gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } static void lowest ( ) { int common_factor = gcd ( num3 , den3 ) ; den3 = den3 / common_factor ; num3 = num3 / common_factor ; } static void addFraction ( int num1 , int den1 , int num2 , int den2 ) { den3 = gcd ( den1 , den2 ) ; den3 = ( den1 * den2 ) / den3 ; num3 = ( num1 ) * ( den3 / den1 ) + ( num2 ) * ( den3 / den2 ) ; lowest ( ) ; } public static void main ( String [ ] args ) { int num1 = 1 , den1 = 500 , num2 = 2 , den2 = 1500 ; addFraction ( num1 , den1 , num2 , den2 ) ; System . out . printf ( "%d/%d + %d/%d is equal to %d/%dNEW_LINE", num1, den1, num2, den2, num3, den3); } }
Largest Divisor of a Number not divisible by a perfect square | Efficient Java Program to find the largest divisor not divisible by any perfect square greater than 1 ; Function to find the largest divisor not divisible by any perfect square greater than 1 ; If the number is divisible by i * i , then remove one i ; Now all squares are removed from n ; Driver Code
public class GFG { static int findLargestDivisor ( int n ) { for ( int i = 2 ; i < Math . sqrt ( n ) + 1 ; i ++ ) { while ( n % ( i * i ) == 0 ) { n = n / i ; } } return n ; } public static void main ( String args [ ] ) { int n = 12 ; System . out . println ( findLargestDivisor ( n ) ) ; n = 97 ; System . out . println ( findLargestDivisor ( n ) ) ; } }
Arithmetic Progression | Java program to check if a given array can form arithmetic progression ; Returns true if a permutation of arr [ 0. . n - 1 ] can form arithmetic progression ; Sort array ; After sorting , difference between consecutive elements must be same . ; driver code
import java . util . Arrays ; class GFG { static boolean checkIsAP ( int arr [ ] , int n ) { if ( n == 1 ) return true ; Arrays . sort ( arr ) ; int d = arr [ 1 ] - arr [ 0 ] ; for ( int i = 2 ; i < n ; i ++ ) if ( arr [ i ] - arr [ i - 1 ] != d ) return false ; return true ; } public static void main ( String [ ] args ) { int arr [ ] = { 20 , 15 , 5 , 0 , 10 } ; int n = arr . length ; if ( checkIsAP ( arr , n ) ) System . out . println ( " Yes " ) ; else System . out . println ( " No " ) ; } }
Check if a number is Triperfect Number | Java code to check if a given number is Triperfect or not ; Returns true if n is Triperfect ; To store sum of divisors . Adding 1 and n since they are divisors of n . ; Find all divisors and add them ; If sum of divisors is equal to 3 * n , then n is a Triperfect number ; Driver program
public class GFG { static boolean isTriPerfect ( int n ) { int sum = 1 + n ; int i = 2 ; while ( i * i <= n ) { if ( n % i == 0 ) { if ( n / i == i ) sum = sum + i ; else sum = sum + i + n / i ; } i += 1 ; } if ( sum == 3 * n & n != 1 ) return true ; else return false ; } public static void main ( String [ ] args ) { int n = 120 ; if ( isTriPerfect ( n ) ) System . out . println ( n + " ▁ is ▁ a ▁ Triperfect ▁ number " ) ; } }
Sum of first N natural numbers which are divisible by X or Y | Java program to find sum of numbers from 1 to N which are divisible by X or Y ; Function to calculate the sum of numbers divisible by X or Y ; Driver code
public class GFG { static int sum ( int N , int X , int Y ) { int S1 , S2 , S3 ; S1 = ( ( N / X ) ) * ( 2 * X + ( N / X - 1 ) * X ) / 2 ; S2 = ( ( N / Y ) ) * ( 2 * Y + ( N / Y - 1 ) * Y ) / 2 ; S3 = ( ( N / ( X * Y ) ) ) * ( 2 * ( X * Y ) + ( N / ( X * Y ) - 1 ) * ( X * Y ) ) / 2 ; return S1 + S2 - S3 ; } public static void main ( String [ ] args ) { int N = 14 ; int X = 3 , Y = 5 ; System . out . println ( sum ( N , X , Y ) ) ; } }
Count numbers from range whose prime factors are only 2 and 3 | Java program to count the numbers within a range whose prime factors are only 2 and 3 ; Function to count the number within a range whose prime factors are only 2 and 3 ; Start with 2 so that 1 doesn 't get counted ; While num is divisible by 2 , divide it by 2 ; While num is divisible by 3 , divide it by 3 ; If num got reduced to 1 then it has only 2 and 3 as prime factors ; Driver code
import java . io . * ; class GFG { static int findTwoThreePrime ( int l , int r ) { if ( l == 1 ) l ++ ; int count = 0 ; for ( int i = l ; i <= r ; i ++ ) { int num = i ; while ( num % 2 == 0 ) num /= 2 ; while ( num % 3 == 0 ) num /= 3 ; if ( num == 1 ) count ++ ; } return count ; } public static void main ( String [ ] args ) { int l = 1 , r = 10 ; System . out . println ( findTwoThreePrime ( l , r ) ) ; } }
Maximum number with same digit factorial product | Java implementation of the approach ; Function to return the required number ; Count the frequency of each digit ; 4 ! can be expressed as 2 ! * 2 ! * 3 ! ; 6 ! can be expressed as 5 ! * 3 ! ; 8 ! can be expressed as 7 ! * 2 ! * 2 ! * 2 ! ; 9 ! can be expressed as 7 ! * 3 ! * 3 ! * 2 ! ; To store the required number ; If number has only either 1 and 0 as its digits ; Generate the greatest number possible ; Driver code
import java . io . * ; class GFG { static String getNumber ( String s ) { int number_of_digits = s . length ( ) ; int freq [ ] = new int [ 10 ] ; for ( int i = 0 ; i < number_of_digits ; i ++ ) { if ( s . charAt ( i ) == '1' || s . charAt ( i ) == '2' || s . charAt ( i ) == '3' || s . charAt ( i ) == '5' || s . charAt ( i ) == '7' ) { freq [ s . charAt ( i ) - 48 ] += 1 ; } if ( s . charAt ( i ) == '4' ) { freq [ 2 ] += 2 ; freq [ 3 ] ++ ; } if ( s . charAt ( i ) == '6' ) { freq [ 5 ] ++ ; freq [ 3 ] ++ ; } if ( s . charAt ( i ) == '8' ) { freq [ 7 ] ++ ; freq [ 2 ] += 3 ; } if ( s . charAt ( i ) == '9' ) { freq [ 7 ] ++ ; freq [ 3 ] += 2 ; freq [ 2 ] ++ ; } } String t = " " ; if ( freq [ 1 ] == number_of_digits || freq [ 0 ] == number_of_digits || ( freq [ 0 ] + freq [ 1 ] ) == number_of_digits ) { return s ; } else { for ( int i = 9 ; i >= 2 ; i -- ) { int ctr = freq [ i ] ; while ( ( ctr -- ) > 0 ) { t += ( char ) ( i + 48 ) ; } } return t ; } } public static void main ( String [ ] args ) { String s = "1280" ; System . out . println ( getNumber ( s ) ) ; } }
Program to find first N Iccanobif Numbers | Java program to find first N Icanobif numbers ; Iterative function to reverse digits of num ; Function to print first N Icanobif Numbers ; Initialize first , second numbers ; Print first two numbers ; Reversing digit of previous two terms and adding them ; Driver Code
public class GFG { static int reversDigits ( int num ) { int rev_num = 0 ; while ( num > 0 ) { rev_num = rev_num * 10 + num % 10 ; num = num / 10 ; } return rev_num ; } static void icanobifNumbers ( int N ) { int first = 0 , second = 1 ; if ( N == 1 ) System . out . print ( first ) ; else if ( N == 2 ) System . out . print ( first + " ▁ " + second ) ; else { System . out . print ( first + " ▁ " + second + " ▁ " ) ; for ( int i = 3 ; i <= N ; i ++ ) { int x = reversDigits ( first ) ; int y = reversDigits ( second ) ; System . out . print ( x + y + " ▁ " ) ; int temp = second ; second = x + y ; first = temp ; } } } public static void main ( String [ ] args ) { int N = 12 ; icanobifNumbers ( N ) ; } }
Add N digits to A such that it is divisible by B after each addition | Java implementation of the approach ; Try all digits from ( 0 to 9 ) ; Fails in the first move itself ; Add ( n - 1 ) 0 's ; Driver Program to test above function
import java . io . * ; class GFG { static int addNDigits ( int a , int b , int n ) { int num = a ; for ( int i = 0 ; i <= 9 ; i ++ ) { int tmp = a * 10 + i ; if ( tmp % b == 0 ) { a = tmp ; break ; } } if ( num == a ) return - 1 ; for ( int j = 0 ; j < n - 1 ; j ++ ) a *= 10 ; return a ; } public static void main ( String [ ] args ) { int a = 5 , b = 3 , n = 3 ; System . out . print ( addNDigits ( a , b , n ) ) ; } }
Count number of triplets ( a , b , c ) such that a ^ 2 + b ^ 2 = c ^ 2 and 1 <= a <= b <= c <= n | Java program to Find number of Triplets 1 <= a <= b <= c <= n , Such that a ^ 2 + b ^ 2 = c ^ 2 ; function to ind number of Triplets 1 <= a <= b <= c <= n , Such that a ^ 2 + b ^ 2 = c ^ 2 ; to store required answer ; run nested loops for first two numbers . ; third number ; check if third number is perfect square and less than n ; Driver code ; function call
class Solution { static int Triplets ( int n ) { int ans = 0 ; for ( int i = 1 ; i <= n ; ++ i ) { for ( int j = i ; j <= n ; ++ j ) { int x = i * i + j * j ; int y = ( int ) Math . sqrt ( x ) ; if ( y * y == x && y <= n ) ++ ans ; } } return ans ; } public static void main ( String args [ ] ) { int n = 10 ; System . out . println ( Triplets ( n ) ) ; } }
Sum of the digits of a number N written in all bases from 2 to N / 2 | Java implementation of the approach ; Function to calculate the sum of the digits of n in the given base ; Sum of digits ; Digit of n in the given base ; Add the digit ; Function to calculate the sum of digits of n in bases from 2 to n / 2 ; to store digit sum in all bases ; function call for multiple bases ; Driver program
import java . io . * ; class GFG { static int solve ( int n , int base ) { int sum = 0 ; while ( n > 0 ) { int remainder = n % base ; sum += remainder ; n = n / base ; } return sum ; } static void SumsOfDigits ( int n ) { int sum = 0 ; for ( int base = 2 ; base <= n / 2 ; ++ base ) sum += solve ( n , base ) ; System . out . println ( sum ) ; } public static void main ( String [ ] args ) { int n = 8 ; SumsOfDigits ( n ) ; } }
Largest number in an array that is not a perfect cube | Java program to find the largest non - perfect cube number among n numbers ; Function to check if a number is perfect cube number or not ; takes the sqrt of the number ; checks if it is a perfect cube number ; Function to find the largest non perfect cube number in the array ; stores the maximum of all perfect cube numbers ; Traverse all elements in the array ; store the maximum if current element is a non perfect cube ; Driver Code
import java . io . * ; class GFG { static boolean checkPerfectcube ( int n ) { int d = ( int ) Math . cbrt ( n ) ; if ( d * d * d == n ) return true ; return false ; } static int largestNonPerfectcubeNumber ( int [ ] a , int n ) { int maxi = - 1 ; for ( int i = 0 ; i < n ; i ++ ) { if ( ! checkPerfectcube ( a [ i ] ) ) maxi = Math . max ( a [ i ] , maxi ) ; } return maxi ; } public static void main ( String [ ] args ) { int a [ ] = { 16 , 64 , 25 , 2 , 3 , 10 } ; int n = a . length ; System . out . print ( largestNonPerfectcubeNumber ( a , n ) ) ; } }
Check if N can be represented as sum of integers chosen from set { A , B } | Java program to find if number N can be represented as sum of a ' s ▁ and ▁ b ' s ; Function to find if number N can be represented as sum of a ' s ▁ and ▁ b ' s ; base condition ; if x is already visited ; set x as possible ; recursive call ; Driver program
import java . util . * ; class solution { static void checkIfPossibleRec ( int x , int a , int b , boolean isPossible [ ] , int n ) { if ( x > n ) return ; if ( isPossible [ x ] ) return ; isPossible [ x ] = true ; checkIfPossibleRec ( x + a , a , b , isPossible , n ) ; checkIfPossibleRec ( x + b , a , b , isPossible , n ) ; } static boolean checkPossible ( int n , int a , int b ) { boolean isPossible [ ] = new boolean [ n + 1 ] ; for ( int i = 0 ; i <= n ; i ++ ) isPossible [ i ] = false ; checkIfPossibleRec ( 0 , a , b , isPossible , n ) ; return isPossible [ n ] ; } public static void main ( String args [ ] ) { int a = 3 , b = 7 , n = 8 ; if ( checkPossible ( a , b , n ) ) System . out . print ( " Yes " ) ; else System . out . print ( " No " ) ; } }
Sum of all odd natural numbers in range L and R | Java program to print the sum of all numbers in range L and R ; Function to return the sum of all odd natural numbers ; Function to return the sum of all odd numbers in range L and R ; Driver Code
import java . io . * ; class GFG { static int sumOdd ( int n ) { int terms = ( n + 1 ) / 2 ; int sum = terms * terms ; return sum ; } static int suminRange ( int l , int r ) { return sumOdd ( r ) - sumOdd ( l - 1 ) ; } public static void main ( String [ ] args ) { int l = 2 , r = 5 ; System . out . print ( " Sum ▁ of ▁ odd ▁ natural ▁ numbers ▁ from ▁ L ▁ to ▁ R ▁ is ▁ " + suminRange ( l , r ) ) ; } }
Sum of common divisors of two numbers A and B | Java implementation of above approach ; Function to calculate gcd of two numbers ; Function to calculate all common divisors of two given numbers a , b -- > input integer numbers ; find gcd of a , b ; Find the sum of divisors of n . ; if ' i ' is factor of n ; check if divisors are equal ; Driver program to run the case
import java . io . * ; class GFG { static int gcd ( int a , int b ) { if ( a == 0 ) return b ; return gcd ( b % a , a ) ; } static int sumcommDiv ( int a , int b ) { int n = gcd ( a , b ) ; int sum = 0 ; for ( int i = 1 ; i <= Math . sqrt ( n ) ; i ++ ) { if ( n % i == 0 ) { if ( n / i == i ) sum += i ; else sum += ( n / i ) + i ; } } return sum ; } public static void main ( String [ ] args ) { int a = 10 , b = 15 ; System . out . println ( " Sum ▁ = ▁ " + sumcommDiv ( a , b ) ) ; } }
Check if a number is formed by Concatenation of 1 , 14 or 144 only | Java program to check if a number is formed by Concatenation of 1 , 14 or 144 only ; Function to check if a number is formed by Concatenation of 1 , 14 or 144 only ; check for each possible digit if given number consist other then 1 , 14 , 144 print NO else print YES ; Driver Code
import java . io . * ; class GFG { static String checkNumber ( int N ) { int temp = N ; while ( temp > 0 ) { if ( temp % 1000 == 144 ) temp /= 1000 ; else if ( temp % 100 == 14 ) temp /= 100 ; else if ( temp % 10 == 1 ) temp /= 10 ; else { return " NO " ; } } return " YES " ; } public static void main ( String [ ] args ) { int N = 1414 ; System . out . println ( checkNumber ( N ) ) ; } }
Fibonacci problem ( Value of Fib ( N ) * Fib ( N ) | Java implementation of the approach ; Driver code
import java . io . * ; class GFG { static int getResult ( int n ) { if ( ( n & 1 ) > 0 ) return 1 ; return - 1 ; } public static void main ( String [ ] args ) { int n = 3 ; System . out . println ( getResult ( n ) ) ; } }