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# codevita,concatenating, primes,concatinating primes

 0 Guys i really need help on this one. Codevita 2k17 just got finished so as u guys know TCS don.t provide solution of their Questions. Here is the question:- If you like numbers, you may have been fascinated by prime numbers. Sometimes we obtain by concatenating two primes. For example, concatenating 2 and 3, we obtain the prime 23. The aim is to find all such distinct "concatenated primes" that could be obtained by concatenating primes ≤ a given integer N. Input Format: Integer N Output Format: M, the number of distinct primes that could be obtained by concatenating two primes ≤ N Constraints: N ≤ 70 Example 1 Input 10 Output 4 Explanations The primes ≤ 10 are 2, 3, 5, 7. These can be used to form the following concatenated numbers: 22, 23, 25, 27, 32, 33, 35, 37, 52, 53, 55, 57, 72, 73, 75, 77. Of these, there are four primes: 23 37 53 and 73. Hence the output is 4. Example 2 Input 20 Output 17 Explanation The prime numbers up to 20 are 2 3 5 7 11 13 17 and 19. Concatenating these two at a time in all possible ways, we get the following numbers: 22 23 25 27 211 213 217 219 32 33 35 37 311 313 317 319 52 53 55 57 511 513 517 519 72 73 75 77 711 713 717 719 112 113 115 117 1111 1113 1117 1119 132 133 135 137 1311 1313 1317 1319 172 173 175 177 1711 1713 1717 1719 192 193 195 197 1911 1913 1917 1919 We have the following 17 primes numbers in this list: 23 37 53 73 113 137 173 193 197 211 311 313 317 719 1117 1319 1913 Hence the output would be 17. I would really appreciate the solution of this question or if someone pointout error in mine code. Here is mine code: import java.util.Scanner; class symbol { public static void main(String[] args) { Scanner inp = new Scanner(System.in); Integer n = inp.nextInt(); if(n<3){ System.out.println("0"); System.exit(0); } int flag = -1; int len; int count=0,count2=0; String primes[] = {"2", "3", "5", "7", "11", "13", "17", "19", "23", "29", "31", "37", "41", "43", "47", "53", "59", "61", "67"}; for(int i=0;i=Integer.parseInt(primes[i])) flag=i; else if(n

 1 you both are double counting.. for example consider: 317 you're creating 317 twice using {31,7} and {7,31}. like this, you are counting many primes twice. Just use a map to check if you already built a prime or not before incrementing the count variable. answered 29 Jul '17, 19:11 600●8 accept rate: 15% Don't know about @stephen26 's code but there is no doubling in mine code...to verify uncomment "// System.out.print(arr[i] + " ");" from last, u'll see the concatenated primes. (30 Jul '17, 03:15) jj_no10★
 0 I dont know the answer either .. here's my code import java.util.*; class prime { public static boolean isPrime(int n) { for (int f = 2; f <=Math.sqrt(n); f++) { if (isFactor(f, n)) return false; } return true; } private static boolean isFactor(int factor, int number) { return number % factor == 0; } public static void main(String args[]) { int[] prime = new int[70]; int n,k=0,count=0; Scanner sc = new Scanner(System.in); n = sc.nextInt(); for (int i = 2; i<=n; i++) { if (isPrime(i)) { prime[k] = i; k++; //System.out.println(i); } } for(int i=0; i
 0 I try this program in python. My program is b=[] e=[] f=[] x=int(input('Enter the last number:')) for i in range(2,x+1): a=0 for j in range(2,x+1): if i%j==0: a+=1 if j==1 or j==i: a-=1 if a==0: b.append(i) print(b,'is prime number') for i in range(0,len(b)): for j in range(0,len(b)): if len(str(b[i]))==1 and len(str(b[j]))==1: d=(b[i]10)+b[j] e.append(d) elif len(str(b[i]))==1 and len(str(b[j]))>1: d=(b[i]100)+b[j] e.append(d) elif len(str(b[i]))>1 and len(str(b[j]))==1: d=(b[i]10)+b[j] e.append(d) elif len(str(b[i]))>1 and len(str(b[j]))>1: d=(b[i]100)+b[j] e.append(d) for i in range(0,len(e)): c=0 for j in range(2,100): if e[i]%j==0: c+=1 if j==e[i]: c-=1 if c==0: f.append(e[i]) print(len(f)) answered 30 Jul '17, 11:38 1 accept rate: 0%
 0 I try this program in python. My program is b=[] e=[] f=[] x=int(input('Enter the last number:')) for i in range(2,x+1): a=0 for j in range(2,x+1): if i%j==0: a+=1 if j==1 or j==i: a-=1 if a==0: b.append(i) print(b,'is prime number') for i in range(0,len(b)): for j in range(0,len(b)): if len(str(b[i]))==1 and len(str(b[j]))==1: d=(b[i]10)+b[j] e.append(d) elif len(str(b[i]))==1 and len(str(b[j]))>1: d=(b[i]100)+b[j] e.append(d) elif len(str(b[i]))>1 and len(str(b[j]))==1: d=(b[i]10)+b[j] e.append(d) elif len(str(b[i]))>1 and len(str(b[j]))>1: d=(b[i]100)+b[j] e.append(d) for i in range(0,len(e)): c=0 for j in range(2,100): if e[i]%j==0: c+=1 if j==e[i]: c-=1 if c==0: f.append(e[i]) print(len(f)) answered 30 Jul '17, 11:42 1 accept rate: 0%

I SUBMITTED THIS CODE AND IT PASSSES THE PUBLIC TEST CASES BUT NOT PRIVATE TEST CASES PLEASE HELP IF I M WRONG ANYWHERE

# include<math.h>

int prime(int n) {

int i,count=0,k=0;

for(i=1;i<=n;i++) {

if(n%i==0)

{

count++;

if(count==1&&i==n/2)

{

return 1;

break;

}

if(count>2)

break;

}

}

if(count==2)

k=1;

else

k=0;

return k;

}

int bits(int w)

{

int c=0;

while(w>0)

{

w=w/10;

c++;

}

return c;

}

int main()

{

long int c=1,i,j=0,n,call=0,total=0,a[100000],recheck=0,count=0,connect=0,bitcheck=0;

scanf("%ld",&n);

for(i=1;i<=n;i++)

{

call=prime(i);

if(call)

{

a[j]=i;

j++;

total++;

}

}

for(i=0;i<total;i++)

{

for(j=0;j<total;j++)

{

bitcheck=bits(a[j]);

connect=a[i]*pow(10,bitcheck)+a[j];

printf("%ld\n",connect);

recheck=prime(connect);

if(recheck)

{

count++;

}

}

}

printf("%ld\n",count);

return 0; }

This answer is marked "community wiki".

4★blazedxi
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accept rate: 0%

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question asked: 29 Jul '17, 14:53

question was seen: 1,315 times

last updated: 01 Aug '17, 11:51