00-Self Test 2. Prime pair conjecture (20)

Let's define D_{N as: Dn = Pn+1-pN, where Pi is the first prime. Obviously there is D1=1 and for n>1 thereis an even number of D N. "The prime number pairs conjecture" that there are infinitely many pairs of adjacent and the difference is 2 primes.}

Now given any positive integer n (<^{5), calculate the number of prime pairs that do not exceed N to satisfy the conjecture.}

input Format: each test input contains 1 test cases, giving a positive integer n.

output format: the output of each test case occupies one line, not exceeding N of the number of prime pairs that satisfy the conjecture.

Input Sample:

20

Sample output:

4

#include <iostream>  using namespace std;    BOOL IsPrime (unsigned long N)       //The prime number of the Web check method  {          if (n <= 3) {          return n > 1;          }       else if (n% 2 = = 0 | | n% 3 = = 0) {          return false;          }       else{for            (unsigned short i = 5; I * I <= n; i + = 6) {               if (n% i = = 0 | | n% (i + 2) = = 0) {                   return false;                   }                  }           return true;      }  }    int main ()  {      int i,j=0,n,c[100000],count=0;      cin>>n;      for (i=1;i<=n;i++) {          if (IsPrime (i))              c[j++]=i;      }      for (i=j-1;i>=1;i--) {          if ((c[i]-c[i-1)) ==2)              count++;      }      cout<<count<<endl;      return 0;  }

The following is written by oneself, not all points.

#include <iostream>

#include <cmath>
using namespace Std;
int main ()
{
int N;
cin>>n;
int a[100000];
a[0]=2,a[1]=3;
int k=2,num=0;
for (int i=3;i<=n;i++)
{
for (int j=2;j<=int (sqrt (i)); J + +)
{
if (i%j!=0)
{
A[k]=i;
k++;
}
}
}
for (int i=1;i<=k;i++)
{
if (a[i]-a[i-1]==2)
num++;
}
cout<<num<<endl;
return 0;
}

00-Self Test 2. Prime pair conjecture (20)