#include <stdio.h>
#include <stdlib.h> // Judge i Is it prime , You need to determine i Can be (long)sqrt(i) Divide the number within
// if i Divisible by one of the prime numbers , be i Not prime ; otherwise i Prime number int main()
//n=10 ans=4
//n=100 ans=25
//n=1000 ans=168
//n=10000 ans=1229
//n=100000 ans=9592
//n=1000000 ans=78498
//n=10000000 ans=664579
//n=100000000 ans=5761455
long i,j,n,flag,ans=,t=;
long *zhi=(long *) malloc (sizeof(long)*);
for (i=;i<=n;i++)
// if "zhi[t]*zhi[t]==i" establish , be i Not prime , Don't continue to judge
// And greater than i You need to use zhi[t] Judge (j=0;j<t+1;j++)
// This method is better than " When the prime number is greater than (long)sqrt(i) Exit from time " Fast
// and (zhi[t]-1)*(zhi[t]-1)<=i<=n
// When i Less than longint The scope can be realized
if (zhi[t]*zhi[t]==i)
for (j=;j<t;j++)
if (i%zhi[j]==)
if (flag)
for (i=0;i<ans;i++)
printf("%ld ",zhi[i]);
if (zhi[ans]==n)
printf("%ld is a prime\n",n);
printf("%ld is not a prime\n",n);
return ;

seek n(n>=2) The prime number within / Judge whether a number is prime —— Method + More articles on detail optimization

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