#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main()
{
    long int n,k,val,fact=1,coeff=0;

    char ch[24],*num,*excl;

    gets(ch);

    num=strtok(ch," ");
    n=atoi(num);

    excl=strtok(NULL,NULL);
    k=strlen(excl);

    while(coeff*k<n)
    {
        val=n-coeff*k;
        fact=fact*val;
        coeff++;
    }


    if(n%k!=0)
        fact=fact*(n%k);
    else fact=fact*k;

    printf("%ld",fact);

    return 0;
}


Hi! I am getting runtime error(access violation) although the solution seems to work for different set of inputs. Please help me out!

Hello,
Should we multiply by k or n%k if (n - i*k) equals k or n%k ?
Which sample is correct?
Sample 1
8!! = 8 *(8-2)*(8-4)*(8-6)*2

Sample 2
8!! = 8 *(8-2)*(8-4)*(8-6)

С 20 не проходит 15 тест!!!

#include <iostream>
#include <string>
using namespace std;
int main() {
    int num, res;
    string c, b = "!", a = "!";

    cin >> num >> c;
    res = num;
    int sign = 0;
    for (int i = 1; b <= c; i++) {
        b = b + a;
        sign = i;
    }
    if (num%sign != 0) {
        for (int i = 1; i*sign < num; i++) {
            res = res * (num - i*sign);
        }
        res = res *(num%sign);
    }
    else {
        for (int i = 1; i*sign < num; i++) {
            res = res * (num - i*sign);
        }
    }
    cout << res;

    return 0;
}

Problem is input of data! How should i fix it?
Code:
#include <iostream>
#include <cstring>
#include <string>

using namespace std;

int n, k;
string s;

int main() {
 cin >> n; cin >> s;
 int i;
 i=0; k=0;
 while(s[i]=='!') {
    k++;
    i++;
 }

 long long a, temp;

 a=1; i=0;
 if(n%k!=0) {
    while(temp!=n%k) {
       temp=n-i*k;
       a*=temp;
       i++;
    }
 } else
 if(n%k==0) {
    while(temp!=k) {
       temp=n-i*k;
       a*=temp;
       i++;
    }
 }

 cout << a << endl;

 system("pause");
 return 0;
}

Am I making it too complicated ? can anyone tell me what's wrong with the logic ? 9 !! works ok
.............

#include "stdafx.h"
#include <iostream>
#include <string>

using namespace std;
int main()
{
    int n;
    string marks;
    int output = 1;
    cin >> n >> marks;
    int k= marks.length();
    int until = n / k;

    if (n % k != 0) {
        for (int i = 0; i <= until; i++) {
             output *=  (n - i * k);
      }
        output *= (n % k);
    }

    else if (n % k == 0) {
        for (int i = 0; i < k; i++) {
            output *= (n - i * k);
        }
        output *= k;
    }
    cout << output << endl;
    return 0;
}

Edited by author 06.08.2016 07:38

never mind...

Edited by author 28.07.2016 07:32

Could anybody give me this test please? Other tests are welcome as well :)

Edited by author 19.07.2016 15:33

Edited by author 05.07.2016 21:07

import java.util.Scanner;

public class Factorials {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        String k = in.next();
        char a[] = k.toCharArray();
        int length = a.length;
        function(n, length);
        in.close();
    }

    private static void function(int n, int length) {
        int t = n % length;
        int c = (t == 0 ? length : t);
        int sum = n;
        for (int i = 1;; i++) {
            sum = sum * (n - i * length);
            if ((n - i * length) == c) {
                System.out.println(sum);
                break;
            }
        }
    }
}

Тест №8. Какой должен быть ответ?

what is test 10?

public class Factorials {

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String str = in.readLine();
        StringTokenizer st = new StringTokenizer(str, " ");

        int n = Integer.parseInt(st.nextToken());
        int k = st.nextToken().length();

        if (n >= 1 && n <= 10) {
            if (k >= 1 && k <= 20) {
                int limite= n % k, total = n, res= 0, i = 1;
                do {
                    res= (n - (k * i));
                    if (res== 0) {
                        System.out.print(total);
                        return;
                    }
                    total *= res;
                    i++;
                } while (res!= limite);
                System.out.print(total);
            }
        }
    }
}

Edited by author 19.09.2015 06:20

According to input data 9!! = 9(9-1*2)(9-2*2)(9mod2) = 9*7*5*1=315

Edited by author 04.06.2015 18:13

Edited by author 04.06.2015 18:13

why wrong ans test case # ?? Plzzz help
my code:
#include<bits/stdc++.h>
using namespace std;

int main()
{
    int a[12],n,i;
//int n,i;
    float res,sum=0;
    cin>>n;

    for(i=0; i<n; i++)
    {
        cin>>a[i];
        sum+=a[i];
    }
    res=(float)sum/n;

    if(res==5)
        printf("Named\n");
    else if(res>=4.5&&res<5)
        printf("High\n");

    else if(res>3&&res<4.5)
        printf("Common\n");
    else
        printf("None\n");
    return 0;
}

Sorry for my english.
First of all check the value of k: 0 < k < n. If False, then the result is n, otherwise -- use formulas.

Can k be gretter than n if possible what should we do in this case?

Если изменить правило 2(если n делится на k) и вместо k умножить на 1, то все заработает.

Определение 1. n!!…! = n(n−k)(n−2k)…(n mod k), если n не делится на k; n!!…! = n(n−k)(n−2k)…1, если n делится на k (знаков ! в обоих случаях k штук).

it's very simple,especially on java.

Если k>n что будет выводить?