In C++, Midy's theorem

We get the integer value a_num which will store the numerator and p_den which will store the denominator which should be a prime number. The task is to check whether the operation after dividing a_num by p_den proves the midy theorem.

The steps to prove Midy's theorem are-

• The input numerator is a_num and the denominator is p_den, which should always be a prime number.

• Divide numbers. Check for repeated decimal values.

• Store decimal values ​​until they do not repeat.

• Check if the numbers are duplicates even, if so then split them in half

• Add the two numbers. If the output is a string of 9's, then it proves Midy's theorem.

Let us see various input and output scenarios for this situation -

In − int a_num = 1 and int p_den = 19

Output− The repeated decimal is: 052631578947368421 Proving Midy’s theorem

Explanation− Follow the above steps to check Midy’s theorem, that is,

• Division 1 / 19 = 052631578947368421

• The repeated decimal value is -: 052631578947368421.

• Cut the number in half, which is 052631578 947368421.

• Add the two halves, which is 052631578 947368421 = 999,999,999.

• As we can see, 999,999,999 is a string of 9, which proves Midi’s theorem.

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Input −int a_num = 49, int p_den = 7

Output − No repeating decimals

Explanation− As we can see, 49/7 does not generate a decimal value because 49 is perfectly divisible by 7. Therefore, the output is "no repeating decimals".

The method used in the following program is as follows

• Enter integer values ​​as int a_num and int p_den.

• Call the function as Midys_theorem(a_num, p_den) to prove Midy’s theorem.

• In the function check_Midys()

  • Create a variable for int first to 0, int to 0 at the end

  • Check whether the function check(val) returns FALSE, and then print Midy's theorem does not apply.

  • ELSE IF len % 2 = 0 Then start looping FOR from i to 0 until i is less than len/2 and set first to first * 10 (str[i] - '0' ) and set last to last * 10 (str[len / 2 i] - '0') and print the proved Midy theorem.

• ELSE, print Midy’s theorem is not applicable.

• Inside the function Midys_theorem(int a_num, int p_den)

  • Create a map type variable to map the integer type value to map_val and clear the map.

  • Set reminder to a_num % p_den.

  • Start equal to 0 when there is no reminder and map_val.find(reminder) is equal to map_val.end() and then set map_val[reminder] to result.length(), reminder to reminder * 10 , temp is reminder/p_den, result is result to_string(temp) and reminder % p_den.

  • Check IF remainder = 0, then return -1 ELSE, set count to result.substr(map_val[reminder])

  • Return count

• Function internal bool check(int val)

  • Loop FOR from i to 2 until i is less than val/2. Checks IF val % i = 0 and returns FALSE, otherwise TRUE.

Example

#include <bits/stdc++.h>
using namespace std;
bool check(int val){
   for(int i = 2; i <= val / 2; i++){
      if(val % i == 0){
         return false;
      }
   }
   return true;
}
void check_Midys(string str, int val){
   int len = str.length();
   int first = 0;
   int last = 0;

   if(!check(val)){
      cout<<"\nNot applicable for Midy&#39;s theorem";
   }
   else if(len % 2 == 0){
      for(int i = 0; i < len / 2; i++){
         first = first * 10 + (str[i] - &#39;0&#39;);
         last = last * 10 + (str[len / 2 + i] - &#39;0&#39;);
      }
      cout<<"\nProved Midy&#39;s theorem";
   }
   else{
      cout<<"\nNot applicable for Midy&#39;s theorem";
   }
}
string Midys_theorem(int a_num, int p_den){
   string result;
   map<int, int> map_val;
   map_val.clear();

   int reminder = a_num % p_den;

   while((reminder != 0) && (map_val.find(reminder) == map_val.end())){
      map_val[reminder] = result.length();
      reminder = reminder * 10;
      int temp = reminder / p_den;
      result += to_string(temp);
      reminder = reminder % p_den;
   }
   if(reminder == 0){
      return "-1";
   }
   else{
      string count = result.substr(map_val[reminder]);
      return count;
   }
}
int main(){
   int a_num = 1;
   int p_den = 19;
   string result = Midys_theorem(a_num, p_den);
   if(result == "-1"){
      cout<<"No Repeating Decimal";
   }
   else{
      cout<<"Repeating decimals are: "<<result;
      check_Midys(result, p_den);
   }
   return 0;
}

Output

If we run the above code it will generate the following output

Repeating decimals are: 052631578947368421
Proved Midy's theorem

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