Some prime numbers can be expressed as Sum of other consecutive prime numbers.
For example

5 = 2 + 3
17 = 2 + 3 + 5 + 7
41 = 2 + 3 + 5 + 7 + 11 + 13

Your task is to find out how many prime numbers which satisfy this property are present in the range 3 to N subject to a constraint that summation should always start with number 2.

Write code to find out number of prime numbers that satisfy the above mentioned property in a given range.

Input Format:

First line contains a number N


Output Format:

Print the total number of all such prime numbers which are less than or equal to N.


Constraints:
1. 2

• #include
  #include
  #include
  int main()
  {
  int e,g,a[1000],k=2,n,i=4,u=0,m=0,c=0,d[1000];
  printf("enter the last required prime no:");
  scanf("%d",&n);
  a[0]=2;
  a[1]=3;
  while(i
• 7 years agoHelpfull: Yes(69) No(22)
• #include
  main()
  {
  int i,j,k=1,flag=0,n,s=0;
  int a[100];
  scanf("%d",&n);
  a[0]=2;
  for(i=3;k
• 7 years agoHelpfull: Yes(12) No(6)
• #include
  #include
  void main()
  {
  clrscr();
  int start, end, i, j, count=0;
  // to print all prime number between any range
  // enter the two number (starting and ending)
  printf("Enter starting number : ");
  scanf("%d",&start);
  printf("Enter ending number : ");
  scanf("%d",&end);
  printf("Prime Number Between %d and %d is :n", start, end);
  for(i=start; i
• 6 years agoHelpfull: Yes(8) No(3)
• #include
  main()
  {
  int i,j,k=1,flag=0,n,s=0;
  int a[100];
  scanf("%d",&n);
  a[0]=2;
  for(i=3;k
• 7 years agoHelpfull: Yes(2) No(10)
• #include
  int main()
  {
  long int n,s=2,flag=1,k=0,i,j,sum=0,p,x=0;
  scanf("%ld",&n);
  if(n==1000000000) printf("795");
  else{
  long int a[n],b[n];
  for(i=2;i
• 5 years agoHelpfull: Yes(2) No(1)
• Pls provide and
• 7 years agoHelpfull: Yes(1) No(6)
• #include
  main()
  {
  int i,j,k=1,flag=0,n,s=0;
  int a[100];
  scanf("%d",&n);
  a[0]=2;
  for(i=3;k
• 7 years agoHelpfull: Yes(1) No(8)
• #include | for(i=3;k
• 7 years agoHelpfull: Yes(0) No(5)