Big Factorials

data type mistake…i could find it out…

Datatype mistake? :expressionless: Do you think you can store 999! with any datatype in C++? Your code will not work. Try to find a more efficient way to solve this problem.

Note: Here is 999! which contains around 2500 (~102500) digits where long double can only hold a maximum 1.7×10308.

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

ll factorial(int n)
{
    if(n > 1)
        return n * factorial(n - 1);
    else
        return 1;
}

int main()
{
    int n;
    cin>>n;

    ll k = factorial(n);

    stringstream ss;
    ss<<k;
    string s;
    ss>>s;

    if(s.size() <= 4) cout<<s<<endl;
    else cout<<s.substr(s.length() - 4)<<endl;

    return 0;
}

What’s wrong here? I am getting “Wrong Answer” on 4th test case.

1 Like

#include <stdio.h>

int main() {
int n,i,fact=1;
scanf("%d",&n);
for(i=1;i<=n;i++){
fact = fact * i;
}

printf("%d",fact % 10000);
return 0;

}

What is the problem in this code?
It is showing wrong answer in test case #3

/******************************************************************************

                          Online C++ Compiler.
           Code, Compile, Run and Debug C++ program online.

Write your code in this editor and press “Run” button to compile and execute it.

*******************************************************************************/

#include

using namespace std;

int main()
{
int i,j,sum=1,k=0;
cin>>i;
for(j=i;j>=1;j–){
sum*=j;

}
if(sum>1000){
cout<<sum%10000;

}
else{
cout<<sum;
}
return 0;
}

help

Did it first try without help… this was fun. :grin:

I actually had the same problem. I still didn’t figure it out where the problem actually is but while debugging I saw that the last four digits of factorial of any number greater than 20 is 0000 but my code was not giving this answer thats why i wrote this code and it worked out. But i still didn’t figure out why the factorial of numbers like 23,24 coming wrong in the past code.

This is my final code:
#include

using namespace std;

int main() {
unsigned long long int y;
unsigned long long int p =1;
cin>>y;
if(y>20){cout<<“0”;}
else{
for(int i = 1;i<y+1;i++){ p = p*i;}
int u=p%10000;
cout<< u;
}
return 0;
}
Hope this helps :slight_smile: .

#include<stdio.h>
int main(){
unsigned long int e,p=1,n,a;
scanf("%ld", &n);
if(n>=0){
for(e=1;e<=n;e++)
p=p*e;

}
if(p<1)
printf("%d", p);
else
a=p%10000;
printf("%ld", a);
return 0;
}

why this is not working for case 4

Sir, I have no idea how to print last four digit of factorial in python… I tried from my own but failed a lot of time, give me a hint… here is my code what I wrote

#factorial

import math

n1 = int(input())

print(math.factorial(n1))

@abir123 The following prints the last digit:

N = 1234567
print(N % 10)

That’s a hint I can give you.

#thanks and lot

i = 1

for i in range(1, 10):
i = i + 1
print(“Thanks a lot”)

1 Like

What is the problem of my code???

#include <stdio.h>

unsigned long long int factorial (unsigned long long int n) {
    if (n == 0)
        return 1;
    else
        return n * factorial (n - 1);
}

int main() 
{
    unsigned long long int n;
    scanf ("%llu", &n);
    int k=factorial(n)%10000;
    printf ("%llu", k);
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int n;
int main()
{
cin>>n;
cout<<(n>19?“0000”:n==8?“0320”:n==13?“0800”:to_string(((long long int)tgamma(n+1))%10000))<<endl;
}

why does it output 8799 for 10 on toph whereas it outputs 8800 everywhere else?

#include <stdio.h>

int main()
{
  int i, j, a[4];
  int n, x;
  unsigned long long int f=1;
  scanf("%d", &n);
  for(i=n;i>=1;i--){
    
  f=f*i;  
  }
  if(f>9999){
   for(j=0;j<4;j++){
            x=f%10;
	    f=f/10;
            a[j]=x;
   }
	printf("%d%d%d%d", a[3], a[2], a[1], a[0]);
  }
  else 
  printf("%llu", f);
  
}

What is the problem of this code

#include <stdio.h>

int main()
{
  int i, j, a[4];
  int n, x;
  unsigned long long int f=1;
  scanf("%d", &n);
  for(i=n;i>=1;i--){
    
  f=f*i;  
  }
  if(f>9999){
   for(j=0;j<4;j++){
     x=f%10;
	    f=f/10;
     a[j]=x;
   }
	 if(a[3]!=0){
	  printf("%d%d%d%d", a[3], a[2], a[1], a[0]);
		}
	  if(a[3]==0) {printf("");}
	}
		
  else 
  printf("%llu", f);
  
}

Why wrong answer in test 4

Using long long integers is not the solution.
You need to assign only 4 digits in fact for the next multiplication with i. So, you just have to add fact%=10000 after fact=fact*i.

My code is stuck at testcase 4. I’d like to know what is the test input at testcase 4.