Odds on C ++

On the hub, several topics appeared about the one-liners in different languages ​​that solved simple problems. I decided to publish several algorithms in C / C ++.
So let's go!

1. The Euclidean Algorithm

Recursive algorithm for finding the greatest common divisor.

int GCD(int a,int b) { return b?GCD(b,a%b):a; } 

2. Finding the NOC

Recursive algorithm for finding the least common multiple.

 int LCM(int a,int b) { return a/GCD(a,b) * b; } 

3. Checking the number 2 ^ n

Algorithm for checking the number of degree 2.

 int isPow2(int a) { return !(a&(a-1)); } 

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4. The function of the exchange of two variables

This algorithm works by using the symmetric distinction property that XOR possesses and does not require the presence of a third variable.

 void swap(int *a, int *b) { *a ^= (*b ^= (*a ^= *b)); } 

5. The exponential algorithm

The power of a number in linear time.
End of recursion condition: if the power of a number is 0, then a ^ 0 = 1.

 int pow(int a,int n) { return (!n)?1:a*pow(a,n-1); } 

6. Indian exponentiation algorithm

The power of the number in logarithmic time.

 int powInd(int a,int n) { return (!n)?1:((n&1)?a:1)*powInd(a*a,n/2); } 

7. Factor numbers

The factorial of a nonnegative integer n.
Continuation of recursion condition: factorial is the product of all natural numbers up to n inclusive.
End of recursion condition: if the number is 0, then 0! = 1.

 int fac(int n) { return n?n*fac(n-1):1; } 

8. Sum of digits

The condition for the continuation of recursion: the sum of digits of a number is equal to the last digit plus the sum of digits of a number without the last digit.
End of recursion condition: if the number is 0, then the sum of digits is 0.

 int count(int a) { return (!a)?0:(a%10+count(a/10)); } 

9. Fibonacci numbers

Fibonacci numbers are elements of a numerical sequence in which each successive number is equal to the sum of the two previous numbers.

 int fib(int n) { return (n<=2)?1:(fib(n-1)+fib(n-2)); } 

10. The next Fibonacci number

The function of finding the Fibonacci numbers.

 int fibNext(int &f1,int &f2) { return f1=(f2+=f1)-f1; } 

11. Mersenne numbers

Mersenne numbers - numbers of the form

 int Mersen(int n) { return !(n&(n+1)); } 

12. Min & Max

 int max(int a,int b) { return (a>b)?a:b; } int min(int a,int b) { return (a>b)?b:a; } 

13. Comparing two numbers

The function returns the difference between two numbers, so if the difference is greater than 0, then the number a is greater than b, if it is equal to 0, then the numbers are the same, otherwise the number a is smaller than b.

 template <typename TYPE> int compare (const TYPE a, const TYPE b){ return ( a - b ); } 

14. Raising the number 2 to the power n

With the help of the shift of unit by n bits, we calculate the two to the power n.

 int pow2(int n) { return 1<<n; } 

One-liners from comments

Finding the NOC from Lertmind

 int lcm(int a, int b) { return (b < 1 ? (b ? lcm(b, a % b) : a) : (a / -lcm(-b, -a % b) * b)); } 

The number of nonzero bits in the number of Mrrl

 int NBit(unsigned int x){ return x==0 ? 0 : (x&1)+NBit(x>>1); } 

The maximum degree of two, which divides n from Mrrl

 int MaxDivPow2(int n){ return n&-n; } 

Compare two numbers from Lertmind

 int cmp(int a, int b) { return (a < b ? -1 : (a > b ? 1 : 0)); } 

or pattern

 template<typename T> int cmp(const T &a, const T &b) { return (a < b ? -1 : a > b); } 

Find the k-th bit in the int * array (assuming sizeof (int) == 4) from Mrrl

 int GetBit(int *a,int k){ return (a[k>>5]>>(k&31))&1; } 

Waiting for more algorithms in comments ...