Some coding template to solve your code easily

Hi, Here I am going to share some of my ready made code that can you use when you are on any competition.
This codes can help you to solve the problem quickly as you need not to type this code as you can copy and paste.

Any numeric system converter

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// this code will help you to convert decimals number to any base numeric system/ // pass the value of 'i' that you need to convert the numeric system. // like if you pass 'i'=2 this code you act as decimal to binary converter. // if you pass 'i'=8 this code will act as decimal to octal converter. // pass the value of 'nm' as the number to convert // here the type of the passing number is as 'long long'. You can change the type as you need. // the converted number is stored in 'dec' and 'dec; is returning the converted value. long long conveter(int i, long long nm) { long long dec = 0, rem, num, base = 1; num = nm; while (num > 0) { rem = num % 10; dec = dec + rem * base; base = base * i; num = num / 10; } return dec; }

Prime Number Checker

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// Returns '0' if its prime, and a '1' if its not prime. int primecker(long long num) { int isprime = 0; for(Long long i = 2; i <= sqrt(num); i ++) { if((num% i) == 0) { isprime = 1; break; } } return isprime; }

Decimal to Binary converter

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// this code will convert decimal number to binary number // just pass an 'int' number and the code will convert the number into binary and will save into 'bin' variable int bin=0; // binary value will store here void dictobin(int n) { if(n/2!=0) dictobin(n/2); bin = bin*10 + n%2; }

Coding Problem' Solution

Problem: The 3n + 1 problem

The 3n + 1 problem

Background

Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all possible inputs.

The Problem

Consider the following algorithm:

 
  1.    input n

  2.    print n



  3.    if n = 1 then STOP



  4.       if n is odd then   



  5.       else   



  6.    GOTO 2

Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than this.)
Given an input n, it is possible to determine the number of numbers printed (including the 1). For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16.
For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.

The Input

The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0.
You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j.
You can assume that no operation overflows a 32-bit integer.

The Output

For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one line and with one line of output for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line).

Sample Input

1 10
100 200
201 210
900 1000

Sample Output

1 10 20
100 200 125
201 210 89
900 1000 174

Solution: C++ 

#include<iostream>
#include<stdio.h>
using namespace std;

int cycle(int num)
{

    int n=num;
    int count=0;
    for(;;)
    {

        count++;
        if(n==1) break;
        else if(n%2!=0)
        {
            n=3*n+1;
        }
        else
        {
            n=n/2;
        }
    }
    return count;

}

main()
{
    int num1, num2;

    while(scanf("%d %d", &num1,&num2)!=EOF)
    {
        int temp=num1;
        int temp2=num2;
        if(num1>num2)
        {
            temp=num1;
            num1=num2;
            num2=temp;

        }
        int maxCycle=0;
        for(; num1<=num2; num1++)
        {

            int totalCycle=cycle(num1);

            if(maxCycle<totalCycle) maxCycle=totalCycle;

        }


        cout<<temp<<" "<<temp2<<" " <<maxCycle<<endl;

    }

}