Skip to main content

Codeforces Round #382 (Div. 2) (D. Taxes)

Goldbach's conjecture is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states that every even natural number greater than 2 is the sum of two prime numbers.

The conjecture has been shown to hold for all integers less than 4 × 1018

Bismillahir Rahmanir Rahim

///Author: Tanvir Ahmmad
///CSE,Islamic University,Bangladesh

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<iterator>
#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array
/*every external angle sum=360 degree
  angle find using polygon hand(n) ((n-2)*180)/n*/
///Floor[Log(b)  N] + 1 = the number of digits when any number is represented in base b
using namespace std;
typedef long long ll;

bool prime(ll num)
{
    for(ll i=2;i*i<=num;i++)
        if(num%i==0) return false;
    return true;
}

int main()
{
    ll num;
    cin>>num;
    if(prime(num)==true) cout<<"1"<<endl;
    else if(num%2==1)
    {
        if(prime(num-2)==false) cout<<"3"<<endl;
        else cout<<"2"<<endl;
    }
    else cout<<"2"<<endl;
    return 0;
}
///Alhamdulillah

Problem Link: https://codeforces.com/problemset/problem/735/D
Labels:

Comments

Post a Comment

Popular posts from this blog

Codeforces Beta Round #53 (A. Square Earth?)

Bismillahir Rahmanir Rahim ///Author: Tanvir Ahmmad ///CSE,Islamic University,Bangladesh #include< iostream > #include< cstdio > #include< algorithm > #include< string > #include< cstring > #include< sstream > #include< cmath > #include< cstring > #include< vector > #include< queue > #include< map > #include< set > #include< stack > #include< vector > #include< iterator > #include   < functional >   ///sort(arr,arr+n,greater<int>()) for decrement of array /*every external angle sum=360 degree   angle find using polygon hand(n) ((n-2)*180)/n*/ ///Floor[Log(b)  N] + 1 = the number of digits when any number is represented in base b using   namespace  std; typedef   long   long ...
Post a Comment
Read more

Codeforces round 1661(B. Getting Zero)

using   second operation  the answer is maximum is  15  because  2 15 =32768 so we need to  pre-calculate all the illegible input 0 to 32768   using  second operation Then  just increment 1(using first operation)  from input check is it  minimum is not ! ///Bismillahir Rahmanir Rahim ///Author: Tanvir Ahmmad ///CSE,Islamic University,Bangladesh #include< iostream > #include< cstdio > #include< algorithm > #include< string > #include< cstring > #include< sstream > #include< cmath > #include< cstring > #include< vector > #include< queue > #include< map > #include< set > #include< stack > #include< vector > #include< iterator > #include   < functional >   ///sort(arr,arr+n,greater<int...
Post a Comment
Read more

Codeforces round 1676(A. Lucky?)

Just count the  first three  &  last three  number ///Bismillahir Rahmanir Rahim ///Author: Tanvir Ahmmad ///CSE,Islamic University,Bangladesh #include< iostream > #include< cstdio > #include< algorithm > #include< string > #include< cstring > #include< sstream > #include< cmath > #include< cstring > #include< vector > #include< queue > #include< map > #include< set > #include< stack > #include< vector > #include< iterator > #include   < functional >   ///sort(arr,arr+n,greater<int>()) for decrement of array /*every external angle sum=360 degree   angle find using polygon hand(n) ((n-2)*180)/n*/ ///Floor[Log(b)  N] + 1 = the number of digits when any number is repres...
Post a Comment
Read more