Inverse thinking to find primes

1#include <stdio.h>2 3 intMainvoid)4 {5     Const intLen = -;6     intPrime[len];7      for(intI=0; i<len; i++)8Prime[i] =1;//1 Mark this ordinal number is a prime, 0 is marked as non-prime9      for(intx=2; x<len; X + +)Ten     { One          for(intj=2; x*j<len; J + +) Aprime[x*j]=0; -     } -      for(intk=2; k<len; k++) the     { -         if(prime[k]==1) -printf"%d", k); -     } +printf"\ n"); -     return 0; +}

I can't read the code, look here:

Assuming a number from 2 to n, I ask for the prime numbers, known as 2 primes. I can assume that all are primes. Using an array prime, such as an element with an ordinal number of 4, it is not a prime, that prime[4]=0.

OK, now create the prime, let all values be initialized to 1, and then let 2 gradually increase the multiples (from twice times to M times, this m*2<=n), these 2 multiples are obviously not prime, you need to change the value of the corresponding angle of the element to 0.

Then 2 increases by 1, repeating the process until N. The remaining angular mark corresponds to an element value of 1, and the value of those angular marks is a prime number. You can print it in turn.

Inverse thinking to find primes