Last non-zero digit of factorial

To find the last non-zero digit of a factorial number, the idea is to remove the extra zeroes in every step multiplying the numbers and don't need to hold the whole big number, just a small digit numbers from the last non-zero digits.
Here goes the code-

#include<stdio.h>
int main()
{
    long long int ans;
        int num;
        scanf("%d",&num);
 
        ans=1;
        for(int i=2;i<=num;i++)
        {
                ans=ans*i;
 
                while(ans%10==0)
                {
                        ans/=10;
                }
                ans%=10000000;
        }
        printf("%lld\n",ans%10);
 
        return 0;
}

Trailing zero in factorial

Finding number of trailing zero in factorial-

#include <stdio.h>
int main()
{
	int n,total,deno;
        scanf("%d",&n);
		
	total=0;
	deno=5;
	while(deno<=n)
	{
		total+=n/deno;
		deno*=5;
	}
	printf("%d\n",total);

	return 0;
}

Sum of Divisors of a Number

Sum of divisors of a Number can be found by the following formula.

Now the code. Following function calculate sum of divisors of a number-

#define i64 long long

i64 power(i64 N,i64 P)
{
	i64 sum=1,i;
	
	if(P==0)return 1;
	else
	{
		for(i=1;i<=P;i++)
			sum=sum*N;
		return sum;
	}
}

i64 sumofdivisor(i64 n)
{
	i64 sum=1,i,count;
	i64 sq=(i64)sqrt(n);
	
	for(i=0;prime[i]<=sq;i++)
	{
		count=0;
		while(n%prime[i]==0)
		{
			count++;
			n/=prime[i];
		}
		sum*=(i64)(power(prime[i],count+1)-1)/(prime[i]-1);
	}
	if(n>1)
		sum*=(n+1);
	return sum;
}

Bitwise sieve

Here is an implementation of Bitwise sieve. For full explanation See this and this-

#define MAX 100000000//max num to prime
#define LMT 10000 //sqrt(MAX)

int flag[MAX/64];
int prime[100000000],total;

#define chkC(n) (flag[n>>6]&(1<<((n>>1)&31)))
#define setC(n) (flag[n>>6]|=(1<<((n>>1)&31)))

void sieve() 
{
	unsigned i,j,k;

	flag[0]|=0;
	for(i=3;i<LMT;i+=2)
		if(!chkC(i))
			for(j=i*i,k=i<<1;j<MAX;j+=k)
				setC(j);

	prime[(j=0)++] = 2;//prime starts from 0 position
	for(i=3;i<MAX;i+=2)
		if(!chkC(i))
		{
			prime[j++] = i;
			//printf("%d\n",primes[j-1]);
		}
	total = j-1;
}

Number of divisor

To find the number of divisors of a number, first calculate find by sieve and then try this-

int number_of_divisor(int num)
{
 int j,count,div=1;
 for(j=0;prime[j]<=sqrt(num);j++) //prime array holds the prime numbers
 {
  count=0;
  while(num%prime[j]==0)
  {
   count++;
   num/=prime[j];
  }
  div*=(count+1);
 }
 if(num>1)
  div<<=1;
 return div;
}

This function returns the number of divisors of the given number.

Number of digits in factorial

This program calculate how many digit in factorial of the number in given base
Anyone can just calculate number of digits in decimal, then b=10 always.

#include <stdio.h>
#include <math.h>
int main()
{
	int number,i,base,digit;
	double c,a=0;

	printf("Give the base and the number: ");
	scanf ("%d%d",&base,&number);

	c=log10(base);
	for (i=number;i>=1;i--) {

		a+=(log10(i))/c; 
	}
	digit=a+1;
	printf ("%d\n",digit);

	return 0;
}

Sieve of Eratosthenes (Prime Finding)

The most popular prime finding method aka Sieve of Eratosthenes.

#define MAX 35000
#define lim 200

bool a[MAX];
int prime[MAX];

void sieve()
{
	int i,j;
	a[0]=1;a[1]=1;          //1 denotes not prime

	for(i=3;i<=lim;i+=2)    //even numbers are exclusive
	{
		if(a[i]==0)     //0 denotes prime
		{
			for(j=i*i;j < MAX;j+=i)
				a[j]=1;
		}
	}

	prime[0]=2;
	j=1;
	for(i=3;i < MAX;i+=2)     //even numbers are exclusive
	{
		if(a[i]==0)
			prime[j++]=i;
	}
}