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如何编写程序以在给定数字后找到 n 个素数？例如，100 后的前 10 个素数，或 1000 后的前 25 个素数。编辑：以下是我尝试过的。我以这种方式获得输出，但我们可以在不使用任何素数测试功能的情况下做到这一点吗？

#include<stdio.h>
#include<conio.h>
int isprime(int);
main()
{
    int count=0,i;
    for(i=100;1<2;i++)
    {
        if(isprime(i))
        {
            printf("%d\n",i);
            count++;
            if(count==5)
                break;
        }
    }
    getch();
}
int isprime(int i)
{
    int c=0,n;
    for(n=1;n<=i/2;n++)
    {
        if(i%n==0)
        c++;
    }
    if(c==1)
        return 1;
    else
        return 0;
}
4

4 回答 4

6

当然。阅读有关Eratosthenes 筛的信息。您无需检查素数，而是生成素数。

于 2012-03-03T04:10:10.933 回答
5

实施埃拉托色尼的“偏移”筛。它只是两个循环，一个接一个，在另一个循环中。

#include <math.h>             // http://ideone.com/38MQlI
#include <stdlib.h>
#include <stdio.h>

typedef unsigned char bool;   // quick'n'dirty

void primes (int n, int above)
{
  double n0 = above / ( log(above) - log(log(above)) ); // ~ 11%..16% overhead
  int i=0, j=0, k=0;
  double top = n*log(above)*log(log(above)) + above;    // approximated
  int lim = sqrt(top), s2 = top - above + 1;

  bool *core = (bool*) calloc( lim+1, sizeof(bool));    // all
  bool *offset = (bool*) calloc( s2+1,  sizeof(bool));  //   zeroes

  for( i=4; i<=lim; i+=2 ) core[i]=1;      // `1` marks composites
  for( i=above%2; i<=s2; i+=2 ) offset[i]=1;            // (even numbers)
  
  for( i=3; i<=lim; i+=2 )
    if( !core[i] )                         // `0` marks primes
    {
      k = 2*i;

      for( j=i*i; j<=lim; j+=k )
          core[j] = 1;

      for( j=(k-(above-i)%k)%k; j<=s2; j+=k )    // hopscotch to the top
          offset[j] = 1;
    }

  printf(" %d +: ",above);
  for( i=0; i<=s2 && n>0 ; ++i )
    if( !offset[i] )                       // not a composite,
    {
      printf(" %d", i);                    // thus, a prime
      --n;
    }
}

int main()
{
  // primes(10,1000);        // 1000 + ... 9 13 19 21 31 33 39 49 51 61
  // primes(10,100000);     // 100000 + ... 3 19 43 49 57 69 103 109 129 151
  primes(10,100000000);    // 100000000 + ... 7 37 39 49 73 81 123 127 193 213
                          // 1000000000 + ... 7 9 21 33 87 93 97 103 123 181
  return 0;
}

您可以在此处添加许多改进。例如，与其标记偶数，不如完全不代表它们。

于 2012-03-04T17:29:50.093 回答
3 
#include <stdio.h>

static int primes[] = {
    2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,
    101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,
    211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,
    307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,
    401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,
    503,509,521,523,541,547,557,563,569,571,577,587,593,599,
    601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,
    701,709,719,727,733,739,743,751,757,761,769,773,787,797,
    809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,
    907,911,919,929,937,941,947,953,967,971,977,983,991,997,
    1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,
    1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,
    1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,
    1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,
    1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499
};
int primeN = sizeof(primes)/sizeof(int);

void printPrime(int n, int count){
    int i;
    for(i=0;primes[i]<n;i++);
    while(count){
        printf("%d\n", primes[i++]);
        count--;
    }
}

int main(){
    printf("first 10 primes after 100\n");
    printPrime(100, 10);
    printf("first 25 primes after 1000\n");
    printPrime(1000, 25);
    getch();
}
于 2012-03-03T13:23:17.370 回答
-1

例如，您想在 100 之后找到 10 个素数。一种方法（不是一种有效的方法）是我们知道 5 个数字是偶数并且不是素数，因此对于其他五个数字，请检查它们的 mod 是否为 (3,5,7,9 ) 如果不是全部为 0，则为素数。

于 2012-03-03T04:11:44.187 回答