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  1. #include<stdio.h>
  2. #include<math.h>
  3.  
  4. int main()
  5. {
  6.     long int t,isprime[100000],i,j,primes[100000],p,sieve[32000];
  7.     for(i=0;i<32000;i++){
  8.             sieve[i]=0;
  9.         }
  10.     j=0;
  11.     for(i=2;i<=(long int)sqrt(1000000000);i++){
  12.             if(sieve[i]==0){
  13.                 primes[j]=i;
  14.                 //printf("%ld\n",primes[j]);
  15.                 j++;
  16.                 p=2;
  17.             while(i*p<=(long int)sqrt(1000000000)){
  18.                 sieve[i*p]=1;
  19.                 p++;
  20.             }
  21.         }
  22.     }
  23.     primes[j]=9999999999;
  24.     //for(i=0;i<=j;i++)
  25.         //printf("%ld\n",primes[i]);
  26.     long int n,m;
  27.     scanf("%ld ",&t);
  28.     while(t--)
  29.     {
  30.         scanf("%ld %ld", &m, &n);
  31.         for(i=0;i<n-m+1;i++){
  32.             isprime[i]=1;
  33.         }
  34.         j=0;
  35.         if(m==1)
  36.             m++;
  37.         while(primes[j]<=(long int)sqrt(n)){
  38.                 //printf("%d\n",primes[j]);
  39.                 i=m/primes[j];
  40.                 if(i==0||i==1)
  41.                     i=2*primes[j];
  42.                 else
  43.                  i=i*primes[j];
  44.                 if(i<m)
  45.                     i+=primes[j];
  46.                 while(i<=n){
  47.                     //printf("%d %d\n",i,primes[j]);
  48.                     isprime[i-m]=0;
  49.                     i+=primes[j];
  50.                 }
  51.                 j++;
  52.         }
  53.         for(i=0;i<n-m+1;i++){
  54.             if(isprime[i]==1){
  55.                 printf("%ld\n",i+m);
  56.             }
  57.         }
  58.         printf("\n");
  59.     }
  60.     return 0;
  61. }
  62.  
  63.  
Success #stdin #stdout 0s 3076KB
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https://ideone.com/PvZdWC
language:
C (gcc 8.3)
created:
9 years ago
visibility:
public

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