Some prime numbers can be expressed as a sum of other consecutive prime numbers. For example 5 = 2 + 3, 17 = 2 + 3 + 5 + 7, 41 = 2 + 3 + 5 + 7 + 11 + 13. Now the 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.
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:
2<N<=12,000,000,000
Sample Input 1 :
20
Sample Output 1 :
2
Explanation :
5 = 2+3
17 = 2+3+5+7
Sample Input 2 :
15
Sample Output 2 :
1
Explanation :
5 = 2+3
Solution: In C
#include <stdio.h>
int prime(int b)
{
int j,cnt;
cnt=1;
for(j=2;j<=b/2;j++)
{
if(b%j==0)
cnt=0;
}
if(cnt==0)
return 1;
else
return 0;
}
int main() {
int i,j,n,cnt,a[25],c,sum=0,count=0,k=0;
scanf("%d",&n);
for(i=2;i<=n;i++)
{
cnt=1;
for(j=2;j<=n/2;j++)
{
if(i%j==0)
cnt=0;
}
if(cnt==1)
{
a[k]=i;
k++;
}
}
for(i=0;i<k;i++)
{
sum=sum+a[i];
c= prime(sum);
if(c==1)
count++;
}
printf("%d",count);
return 0;
}Solution: In Java
import java.util.Scanner;
class Main {
static int prime(int b) {
int j,cnt;
cnt=1;
for (j = 2; j <= b/2; j++) {
if(b%j==0)
cnt=0;
}
if(cnt==0)
return 1;
else
return 0;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int i,j,n=0,cnt,c=0,sum=0,count=0,k=0;
Main t = new Main();
int[] a = new int[25];
n = sc.nextInt();
for (i = 2; i <=n ; i++) {
cnt=1;
for (j = 2; j <= n/2; j++) {
if(i%j==0)
cnt=0;
}
if(cnt==1) {
a[k]=i;
k++;
}
}
for (i = 0; i < k; i++) {
sum=sum+a[i];
c=t.prime(sum);
if(c==1)
count++;
}
System.out.println(count);
}
}Solution: In Python 3
num = int(input())
arr = []
sum = 0
count = 0
if num > 1:
for i in range(2, num + 2):
for j in range(2, i):
if i % j == 0:
break
else:
arr.append(i)
def is_prime(sum):
for i in range(2, (sum // 2) +2):
if sum % i == 0:
return False
return True
for i in range(0, len(arr)):
sum = sum + arr[i]
if sum <= num:
if is_prime(sum):
count = count + 1
print(count)Also Checkout