[Reprint] Basic algorithm--sieve method to calculate prime number

Following from http://blog.csdn.net/stack_queue/article/details/53560887

It is a common problem in the program design competition to find primes, and the most basic method is to judge directly by the definition of prime number, which can only be divided by 1 and it is prime. This method is suitable to determine whether a single number is a prime, when a range of primes is required and this range is relatively large, this method is not used, and even the program to run a few minutes to calculate the results.

The idea of sieve is to remove all composite within the required range, and the remainder is prime, and any composite can be expressed as a product of prime numbers, so if a number is known as prime, its multiples are composite.

Following from http://blog.csdn.net/morewindows/article/details/7347459

The simplest method of sieve prime method is to start from 2, so the multiples of 2 are removed, and then starting from 3, the multiples of 3 are removed. According to this it is easy to write the code, the following code is the number of sieve method to get a prime number within 100 and saved to the primes[] array.

//by Morewindows (http://blog.csdn.net/MoreWindows )  Const intMAXN = -; BOOLFLAG[MAXN]; intPRIMES[MAXN/3], pi; voidgetprime_1 () {intI, J; Pi=0; memset (Flag,false,sizeof(flag));  for(i =2; i < MAXN; i++)          if(!Flag[i]) {Primes[pi++] =i;  for(j = i; J < maxn; J + =i) flag[j]=true; }  }  

It can be seen that there will be a lot of repeated visits, such as access to flag[2] and flag[5] Each visit flag[10] once. It is therefore best to think of ways to reduce this repeated access, so that each element of the flag[] array is only accessed once.

This can be considered-the simple sieve prime method is to use a multiple of a prime number must not be a prime, and the product of any one and all other primes is necessarily not prime (this is because each composite must have a minimum element factor).

In order to experiment with this idea, the number from 2 to 10 is first verified.

2,3,4,5,6,7,8,9,10 the flag is unmarked at the beginning.

First step access 2,flag[2] No tag so add 2 to the Prime number table, and then multiply the 2 with all the numbers in the Prime number table is necessarily not prime, 2*2=4 so Mark Flag[4].

2,3,4, 5,6,7,8,9,10

Step two access 3,flag[3] No tag so add 3 to the Prime number table, multiply the 3 with all the numbers in the Prime number table is necessarily not prime, 3*2=6,3*3=9 so Mark Flag[6] and flag[9].

2,3,4, 5,6, 7,8,9, 10

Step three access 4,flag[4] There is a tag so 4 does not join the Primes table, the number of 4 and all the numbers in the Prime number table is necessarily not a prime, 4*2=8,4*3=12 so Mark Flag[8].

2,3,4, 5,6, 7,8,9, 10

Fourth Step access 5,flag[5] No tag so add 5 to the Prime number table, multiply the 5 with all the numbers in the Prime number table is necessarily not prime, 5*2=10,5*3=15 so Mark Flag[10].

2,3,4, 5,6, 7,8,9,ten

Fifth Step access 6,flag[6] There is a tag so 6 does not join the Primes table, the number of 6 and the number of all numbers in the table is bound to be not prime, 6*2=12,6*3=18,6*5=30.

2,3,4, 5,6, 7,8,9,ten

1 //by Morewindows (http://blog.csdn.net/MoreWindows )  2 Const intMAXN = -; 3 BOOLFLAG[MAXN]; 4 intPRIMES[MAXN/3], pi; 5 voidgetprime_2 ()6 {  7     intI, J; 8PI =0; 9memset (Flag,false,sizeof(flag)); Ten      for(i =2; i < MAXN; i++)   One     {   A         if(!Flag[i]) -primes[pi++] =i;  -          for(j =0; (J < Pi) && (i * primes[j] < MAXN); J + +)   theFlag[i * Primes[j]] =true;  -     }   -}

Is this code right? Careful review of the analysis process, you can find that some of the data has been visited many times, which is certainly not the result we hope, our request is to let each composite only its minimum factor sieve once . For example, 12, its minimum factor is 2, so it should only be in the calculation of 6*2 to access, and should not be in the calculation of 4*3 to access, the same 18 should be in the calculation 9*2 only to access, and should not be in the calculation of 6*3 to access.

Find out why, and then think about how to solve it. 6*3 not 9*2 can be, because 6 is a multiple of 2, so after the calculation of 6*2 can no longer multiply 6 and 2 large primes, the result of these multiplication will inevitably lead to repeated calculations. So for any number, if it is a multiple of that prime, it can no longer be multiplied by the prime number in the prime number table, such as 9 is a multiple of 3, so after 9*3 can no longer be used to calculate 9*5. So add a line of judgment to the code:

1 //by Morewindows (http://blog.csdn.net/MoreWindows )  2 Const intMAXN = -; 3 BOOLFLAG[MAXN]; 4 intPRIMES[MAXN/3], pi; 5 voidgetprime_2 ()6 {  7     intI, J; 8PI =0; 9memset (Flag,false,sizeof(flag)); Ten      for(i =2; i < MAXN; i++)   One     {   A         if(!Flag[i]) -primes[pi++] =i;  -          for(j =0; (J < Pi) && (i * primes[j] < MAXN); J + +)   the         {   -Flag[i * Primes[j]] =true;  -             if(i% primes[j] = =0)//this guarantees that each non-vegetarian number is only screened once. -                  Break;  + }   -     }   +}

[Reprint] Basic algorithm--sieve method to calculate prime number