1. Judgment of prime numbers:

#include<bits/stdc++.h>
typedef long long LL;
const int MaxN = 1e5;
using namespace std;

int a[MaxN + 5],n;
int is_Prime(int x)
{
	if(x==1 || x==2) return 0;
	for(int i=2; i*i<=x; ++i)
		if(x%i == 0)return 0;
	return 1;
}
int main(){
    while(cin>>n){
		if(is_Prime(n))
			cout<<n<<" is Prime"<<endl;
		else cout<<n<<" not is Prime"<<endl;
    }
    return 0;
}

2. Maximum common divisor and minimum common multiple:

#include<bits/stdc++.h>
typedef long long LL;
const int MaxN = 1e5;
using namespace std;

int a[MaxN + 5],m,n;
int main(){
    while(cin>>n>>m){
    	int x = __gcd(m,n);  //Greatest common divisor
    	int y = m * n / x;   //Minimum common multiple
    	printf("%d and %d The greatest common divisor is %d\n",m,n,x);
    	printf("%d and %d The lowest common multiple is %d\n",m,n,y);
    }
    return 0;
}

3. Conversion between any binaries:

#include<bits/stdc++.h>
typedef long long LL;
const int MaxN = 1e5;
using namespace std;

int a[MaxN + 5],m,n;//Converting decimal m to n-ary
stack<int>s;
int main(){
    while(cin>>m>>n){
    	int k = 0;
    	while(m>0){
			int e = m%n;
			s.push(e);
			m /= n;
    	}
    	while(!s.empty()){
			cout<<s.top();
			s.pop();
    	}
    	cout<<endl;
    }
    return 0;
}

4. Implementation of atoi() function

Function of the atoi() function: converting strings to integers

Output upper or lower bounds: -2147483648~2147483647

#include<bits/stdc++.h>
typedef long long LL;
const int MaxN = 1e5;
using namespace std;

int a[MaxN + 5],m,n;
char s1[10],s2[10];
int main(){
    while(cin>>s1>>s2){
		LL x = atoi(s1);
		LL y = atoi(s2);
		cout<<x<<endl<<y<<endl;
    }
    return 0;
}

5. Determine if a number is a perfect square number:

#include<bits/stdc++.h>
typedef long long LL;
const int MaxN = 1e5;
using namespace std;

int a[MaxN + 5],m,n;
int main(){
    while(cin>>n){
    	if(sqrt(n) == (int) sqrt(n))  //sqrt(x) == (int) sqrt(n) ? YES:NO;
    		cout<<n<<" is a Perfect Squares"<<endl;
    	else
    		cout<<n<<" not is a Perfect Squares"<<endl;
    }
    return 0;
}

6. Determine whether a number is a power of 2

When the power of 2 is written in binary form, it is easy to see that there is only one 1 in binary, followed by n zeros.If you subtract one from this number, the only one will become zero, and the original n zeros will become 1; therefore, when you run and run the original number and the number subtracted from one, you will find zero.

For example: 2 (10), 4 (100), 8 (1000), 16 (10000)...

(n & n-1) == 0