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C – Chef and Maths CodeChef Solution

Problem -C – Chef and Maths CodeChef Solution

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C – Chef and Maths CodeChef Solution in C++14

// i am seven star coder 
// i am working in google

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

#define int long long
#define ll long long
#define ld long double
#define f first
#define s second
int mod=1e9+7;



void solve(){
    
int arr[1000001];int first=1;

    arr[0]=1;
for(int i=1; i<1000001; i++){
    first=(i*first)%mod;
arr[i]=(first*arr[i-1])%mod;
}
    int n;
    cin>>n;
    while(n--){
        int s;
        cin>>s;
        cout<<arr[s]<<'\n';
    }
}

int32_t main()
{
    ios_base::sync_with_stdio(false);
        cin.tie(0); cout.tie(0);
    int t=1;
    // cin>>t;
    while(t--){
        solve();
    }
    
    
    return 0;
}

C – Chef and Maths CodeChef Solution in PYTH 3

# cook your dish here
# cook your dish here
t = int(input())
s = []
for i in range(t):
    n = int(input())
    s.append(n)
n = max(s) + 1
l = [0 for x in range(n)]
p = 1
a = 1
for j in range(1, n):
    a = (a * j) % 1000000007
    p = p * a % 1000000007
    l[j] = p
for k in s:
    print(l[k]) 

C – Chef and Maths CodeChef Solution in C

#include <stdio.h>

// loops
#define rep(i, n) for(int i=0; i<n; i++)
#define repA(i, a, n) for(int i=a; i<n; i++)
#define repD(i, a, n) for(int i=a; i>=n; i--)


// container and contants
#define N 1000001

typedef long long int ll;
typedef unsigned long long ull;

// global constants
int mod=1e9+7;
//modular exponentation
ll modpow(int x, int y)
{
    ll res = 1, tmp;
    x = x % mod;

    while (y > 0)
    {
        if (y & 1)
        {
            tmp=res;
            tmp*=x;
            res = tmp % mod;
        }
        y = y>>1;
        tmp=x;
        tmp*=x;
        x = tmp % mod;
    }
    return res;
}
//modular multiplicative inverse
ll modInverse(ll a){
	ll z = mod;
    ll q = 0, p = 1;
    while(a > 1){
        ll s = a/mod;
        ll t = mod;
        mod = a % mod, a = t;
        t = q;
        q = p - s * q;
        p = t;
    }
    if (p < 0)
       p += z;
	mod=z;
    return p;
}
int main()
{
    ll fact[N], ans[N];
    ans[0]=fact[0]=1;
    repA(i, 1, N){
        fact[i]=fact[i-1];
        fact[i]*=i;
        fact[i]%=mod;
        ans[i]=ans[i-1];
        ans[i]*=fact[i];
        ans[i]%=mod;
    }
    int t, n;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        printf("%lld\n", ans[n]);
    }
    return 0;
}

C – Chef and Maths CodeChef Solution in JAVA

import java.io.*;
import java.util.*;

class mathl {
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        long mod = 1000000007;
        long arr[]=new long[1000001];
        long fact[]=new long[1000001];
        fact[0]=1;arr[0]=1;
        for(int i=1;i<=1000000;i++){
            fact[i] = (fact[i-1]*i)%mod;
            arr[i]=(arr[i-1]*fact[i])%mod;
        }
        int t = sc.nextInt();
        for(int z=1;z<=t;z++){
            int n=sc.nextInt();
            pw.println(arr[n]);
        }
        pw.close();
    }
}

C – Chef and Maths CodeChef Solution in PYPY 3

from sys import stdin,stdout
from collections import defaultdict
input=stdin.readline
print=stdout.write
T=int(input())
M=1000000007
n=1000005
Array=[1]
f=1
for j in range(1,n+1):
    f=f*j
    Array.append((f*Array[j-1])%M)
    f=f%M
for i in range(T):
    n=int(input())
    print(str(Array[n])+"\n")

C – Chef and Maths CodeChef Solution in PYTH

MX = 10**6
F = [0 for x in range(MX+1)]
MD = 1000000007
F[1] = 1
n = 1
for k in range(2,MX+1):
	n = n*k%MD
	F[k] = F[k-1]*n%MD
# endfor k
t = int(raw_input())
for i in range(t):
	N = int(raw_input())
	print F[N]
# endfor i
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