Can Anyone be kind enough to explain this ques to me? I'll be grateful to you..

1) 11 !!!! - 231
2) 13 !! - 135135
3) 36 !!!!!! - 33592320
4) 7 !!!!! - 14
5) 21 !!!!!!! - 2058

and DONT forget that (n mod k) or k is not the last multiplayer
rus: не путайте, (n mod k) или k не являются последним множителем!!

//Don't ask me. I don't know. Just work. If anybody can tell me why   for (int i = k + 7; i >= 0; i--)   and why we needn't n%k - Please, tell! I don't understand.

using System;

namespace ConsoleApp32
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] vvod = Console.In.ReadLine().Split(new char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries);
            int n = int.Parse(vvod[0]);
            string k1 = vvod[1];
            int k = k1.Length;
            int q = 1;

            for (int i = k + 7; i >= 0; i--)
                {
                    int t;
                    t = n - i * k;
                    if (t > 0) q *= t;
                }
            Console.WriteLine(q);
        }
    }
}

Edited by author 13.01.2018 23:49

For n = 10 and k = 20, what should be the output? Well, obviously it's not 0!

be attentive with loop (while)! we know formule n - x * k =) and in depends of number parity condition will be >= (ex. [n - x * k >= n mod k] or [n - x * k >= k]). Good luck!)

Edited by author 19.01.2023 16:23

what is wrong with my program? I used Python
code below:

n = input()
list = n.split()
n = int(list[0])
k = len(list[1])

if (n < k):
    p = k
else:
    p = n

i=2
num = n - k
if (n>=1 and n<=10) and (k>=1 and k<=20) and (num > 0):
    if((n%k) == 0):
        while(num>k):
            p = p*num
            num = n-i*k
            i=i+1

        p = p*k
    else:
        while(num>(n%k)):
            p = p*num
            num = n-i*k
            i=i+1
        p = p*(n%k)

print(p)

Edited by author 17.06.2022 20:39

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

int solve(int n, string k){
    int ans=k.size();

    if(n>=1){
        return n*solve(n-ans);

    }
    else{
        return 1;


    }


}



int main()
{
   int  t,n;
    string k;
    cin>>n;
    cin>>k;
    int result=solve(n,k);
    cout<<result<<endl;


    return 0;
}

тест 1: n=9 k=2
тест 2: n=1 k=1
тест 3: n=10 k=1
тест 4: n=8 k=1
тест 5: n=10 k=2
тест 6: n=7 k=2
тест 7: n=1 k=2
тест 8: n=2
тест 9: n=10 k=5
тест 10: n=9 k=5
n=3,4,5,6 не встречается...

#include<bits/stdc++.h>

using namespace std;

int main()
{
    int n;
    int mul = 1;
    string s;
    cin>>n>>s;
    int k = s.size();
    if(n%k==0){
        int in = n;
        int temp = k;
        while(1){
            mul = mul*in;
            in = in - k;
            if(in==k){
                mul*=in;
                break;
            }
        }
    }
    else{
        int in = n;
        while(1){
            mul = mul*in;
            in = in - k;
            if(in==(n%k)){
                mul*=in;
                break;
            }
        }
    }
    cout<<mul<<endl;
    return 0;
}

What is test case 2?

#I don't know why i start from lk+7. If anyone knew about that, please notify me.

import sys

n,k = sys.stdin.readline().split()
n = int(n)
lk = len(k)
res = 1

for i in range(lk+7,-1,-1):
    t = n - i*lk
    if t>0: res *= t

sys.stdout.write("%d\n"%res)

Edited by author 16.07.2021 09:03

what is wrong?

#include <iostream>
#include <cmath>
#include <map>
#include <vector>
using namespace std;

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

    int k = str.size();
    int i = 0;
    int sum = 1;
    if(n % k == 0)
    {
        while(i != n / k)
        {
            sum *= n - k * i;
            i++;
        }
    }
    else
    {
        while(i != n / k + 1)
        {
            sum *= n - k * i;
            i++;
        }
        sum *= n % k;
    }

    cout << sum << endl;

    return 0;
}

Edited by author 31.12.2015 12:38

#include<iostream>
//#include<string>
using namespace std;
int main()
{
    int n;
    string s;
    cin>>n;
    cin>>s;
    int sum=n;
    int len = s.size();
//cout<<len<<endl;

    int i =1;
    while((n-i*len)>1)
    {
        sum*=(n-i*len);
        i++;
    }
        if(n%len!=0)
            cout<<(sum*(n%len))<<endl;
        else
            cout<<sum*len<<endl;

        return 0;
    }

A simple way to solve is to begin with a value x which going to depend whether n % k == 0. and increase it by k until it becomes n. and of course on each step multiply it to your sum which is initially 1.

for n=2,k=3
ans=2*(2-3)*(2-6)*(2-9)*.....
The author must include the condition that n>=k..

На Паскале не работает код если начать его так:

read(n); read(s); (или readln(s))

n - longint, s - string

Но если так:

readln(s);

А затем разрезать на n и s1, то всё ОК.
Разве так и должно быть?

a,b=input().split()
n=int(a)
k=len(b)
temp=n
t=k
while temp-k>0:
        n*=(temp-k)
        k+=t

print("%d"%n)

import java.util.*;
import java.lang.*;
public class Factorial{
        public static void main(String[] args){
                Scanner cin=new Scanner(System.in);
                int n=Integer.parseInt(cin.next());
                String k=cin.nextLine();
                int j=k.length()-1;
                int sum=1;
                for(int i=n;i>0;i-=j){
                        sum=sum*i;
                }
                System.out.println(sum);
        }
}

using namespace std;
#include<iostream>
#include<string>
#include<sstream>
#include<ctype.h>
void factorial(unsigned long n,unsigned long k);
int main()
{
    unsigned long counter=0;
    string s;
    unsigned long n;

    while(getline(cin,s))
    {
    counter=0;
    string::size_type position,position2;
    position=s.find(' ');
    string num;
    for(int i=0;i<position;i++)
    {
        if(isdigit(s[i]))
            num+=s[i];

    }
    stringstream ss(num);
    ss>>n;
    position2=s.find('!');
    if(position2==position+1)
    {
       for(int i=position+1;i<s.size();i++)
        if(s[i]=='!')
        counter++;
   factorial(n,counter);
    }

    }
}

void factorial(unsigned long n,unsigned long k)
{
  unsigned long res=1;
  int mod=n%k;
  unsigned long mul=n;
    if(mod!=0)
    {
        while(mul>1)
        {
            res*=mul;
            mul=mul-k;
        }
        cout<<res<<endl;
    }
    else
    {
        while(mul>=k)
        {
            res*=mul;
            mul=mul-k;
        }
        cout<<res<<endl;
    }
}

Edited by author 05.04.2018 15:46

Edited by author 05.04.2018 15:47

#include <iostream>
#include <string>

using namespace std;

int main()
{
    int n,m=1;
    string k;
    cin>>n;
    cin>>k;
    int j=k.length();
    for(int i=n;i>=(n%j);i-=j){
        m=m*i;

    }


    cout<<m<<k;



    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main(){
int n, sana,s;
char k[21];
cin>>n;
cin.getline(k,sizeof(k));
sana = strlen(k)-1;

if (1<=n&&n<=10&&1<=sana&&sana<=20){

if(n%sana==0) { s=sana;}
else          { s=1;}
while(n>0){
    s*=n;
    n=n-sana;}
cout << s << endl;
}
}

Edited by author 28.01.2018 23:09