【PAT】1007. Prime Pair Conjecture (20)

1007. Prime Pair Conjecture (20)

Let us define d _n  as: d _n  = p _n+1  - p _n , where p _i  is the ith prime number. Clearly there is d ₁ =1 and d _n  is even for n>1. The "prime pair conjecture" states that "there are infinitely many pairs of adjacent prime numbers that differ by 2".

Now given any positive integer N (< 10 ⁵ ), count the number of pairs of prime numbers not exceeding N that satisfy the conjecture.

Input format: Each test input contains 1 test case, giving a positive integer N.

Output format: The output of each test case occupies one line, and does not exceed the number of prime pairs of N satisfying the conjecture.

Input sample:

20

Sample output:

4 

About the prime number judgment function:

　　Type 1: Judging according to the definition, when N (<10 ⁵ ) is large, this kind of function has a very large amount of calculation, and displays "operation timeout" after submitting the program:

1 bool prime(int num){
2     int i;
3     if(num==1) return false;
4     for(i=2;i<num;i++)
5         if(num%i==0) return false;
6     return true;
7 }

　　Type 2: Judging within the range of the square root of N, which greatly reduces the amount of computation, so use this function:

1 bool prime(int num){
2     int i,k;
3     if(num==1) return false;
4     k=int(sqrt(num));    
5     for(i=2;i<=k;i++)
6         if(num%i==0) return false;
7     return true;
8 }

 

programming:

method one:

　　1. Find all prime numbers within the range of N and store them in the array prime_num[]. Since there are a total of 9592 prime numbers within the range of 10 ^{5 , the number of array elements should not be less than 9592}

　　2. Compare the elements in the prime_num[] array to find the number of the last digit that is 2 larger than the previous digit

The C++ code is as follows:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 bool prime(int num){
 4     int i,k;
 5     if(num==1) return false;
 6     k=int(sqrt(num));    
 7     for(i=2;i<=k;i++)
 8         if(num%i==0) return false;
 9     return true;
10 }
11 int main() {
12     int prime_num[9600];
13     int i,j=0,count=0;
14     long n;
15     cin>>n;
16     for(i=2;i<=n;i++){
17         if(prime(i)){
18             prime_num[j]=i;
19             j++;
20         }
21         else continue;
22     }
23     for(i=0;i<j-1;i++)
24         if(prime_num[i+1]-prime_num[i]==2) ++count;
25     cout<<count;
26     system("pause");
27     return 0;
28 }

C++ Code 1

 

Method Two:

　　When N>2, all prime numbers are odd, so only the number of adjacent odd numbers that are all primes is required.

The C++ code is as follows:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 bool prime(int num){
 4     int i,k;
 5     if(num==1) return false;
 6     k=int(sqrt(num));    
 7     for(i=2;i<=k;i++)
 8         if(num%i==0) return false;
 9     return true;
10 }
11 int main() {
12     int i,count=0;
13     long n;
14     cin>>n;
15     for(i=3;i+2<=n;i+=2){
16         if(prime(i))
17             if(prime(i+2)) count++;        
18         else continue;        
19     }
20     cout<<count;
21     system("pause");
22     return 0;
23 }

C++ Code 2