c# - Sum of Numbers as Distinct Primes -

//list style   using system; using system.collections.generic; using system.linq;   public class pr{      static public void main (){              int n, i, j, k, l, sum,flag = 0;             //int sum = i+j;             //int k = (n-i);             //int l = (n-j);              //system.console.writeline ("enter number");             //n = convert.toint32 (console.readline());              //list <int> primes = new list <int>(); //list handle numbers             //hashset <int> myprimes = new hashset <int> (primes);                    system.console.writeline ("enter number");                  n = convert.toint32 (console.readline());                  //myprimes.add(n);                  //myprimes.add(i);                  //myprimes.add(j);                  // var count = string.join(", ", primes);                   //system.console.writeline("the value of n {0}",myprimes);                      for(i=3; i<n/2; i++){                          for(j=3; j<n/2; j++){                              if(checkprime(i) == 1){                                  if(checkprime(j) == 1){                                      if (checkprime(n-i) == 1){                                          if (checkprime(n-j) == 1){                                                  //if(i == j){                                                 //sum = i+j;                                               system.console.writeline("{0}={1}+{2}\n",n,i,n-i);                                         //}                                      }                                 }                             }                         }                              if (flag == 0 && (n-i) <= 0 && (n-j) <= 0){ //check avoid dupes                                      if (n <= 0 && <= 0 && j <= 0){                                          console.write("{0}\n",n);                                     }                               }                           }                     }      }              public static int checkprime(int n){                  int i, j, flag = 1;                  (i = 2; i<=(math.sqrt(n)); i++){                       (j = 2; j<=(math.sqrt(n)); j++){                           if (n%i == 0 && n%j == 0 ){ //even number check                                  i++;                                 j++;                                 flag = 0;                     }                   }              }                  return flag;             }       } 

so have been experimenting while now. cant seem print possible solutions. example 24 able print 7+17 not 2+5+17. there answers being repeated , might have fact dont have duplicate checks. tried push integers in list , use hashset have distinct integers got stuck , tried brute force it. numbers printed supposed distinct prime integers. dont understand how print distinct numbers , there elegant way print out possible.

thanks help!

don't know if it's elegant enough you, i've mashed dirty way make work:

static void main()     {         console.writeline("enter number");         var numbertosum = convert.toint32(console.readline());          var primesinrange = getprimesupto(numbertosum);         var foundsolutions = primesinrange.subsetsof().where(prime => prime.sum() == numbertosum);          foreach (var solution in foundsolutions.tolist())         {             var formatoperation = solution                 .select(x => x.tostring())                 .aggregate((a, n) => + " + " + n) + " = " + numbertosum;              console.writeline(formatoperation);         }          console.readline();     }      public static ienumerable<int> getprimesupto(int end)     {         var primes = new hashset<int>();          (var = 2; <= end; i++)         {             var ok = true;              foreach (var prime in primes)             {                 if (prime * prime > i)                     break;                  if (i % prime == 0)                 {                     ok = false;                     break;                 }             }              if (ok)                 primes.add(i);         }          return primes;     }      public static ienumerable<ienumerable<t>> subsetsof<t>(this ienumerable<t> source)     {         if (!source.any())             return enumerable.repeat(enumerable.empty<t>(), 1);          var element = source.take(1);          var havenots = subsetsof(source.skip(1));         var haves = havenots.select(set => element.concat(set));          return haves.concat(havenots);     } 

i've found solution quite dirty divided problem more understandable. getprimesupto returns prime number 2 number you've provided in input, subsetsof returns combination of numbers summed equals input number you've provided , foreach in main produces formatted output easy on eye. hope helps!