Question

我的算法有问题。我在下面的任务中找不到问题所在。

说明：

<块引用>

10 以下的素数之和为 2 + 3 + 5 + 7 = 17。找出所有小于 200 万的质数之和。

这是我在 C 中的解决方案：

#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>

int main() {
    bool *numbers = (bool *)malloc(sizeof(bool) * 2000000);
    unsigned long run = 1;
    while (run < 2000000) {
        numbers[run] = true;
        run++;
    }
    unsigned long sum = 0;
    for (long i = 2; i < 2000000; i++) {
        if (numbers[i] == true) {
            for (long x = 2 * i; x < 2000000; x += i) {
                numbers[x] = false;
            }
        }
    }
    run = 0;
    while (run < 2000000) {
        if (numbers[run] == true) {
            sum = sum + run; 
        }
        run++;
    }
    printf("%d\n", sum-1); // cause 1
    free(numbers);
    return 0;
}

感谢您的帮助！！！

Answer 1

您代码中的问题是您没有为类型为 printf 的 sum 使用正确的 unsigned long 转换说明符：您应该编写：

printf("%lu\n", sum);

结果，142913828923 超出了 32 位整数的范围，因此您实际上应该使用更大的类型，例如 long long，它保证至少有 63 个值位。最后一个循环应从 2 开始，以避免将 1 计为质数。

这是一个改进的版本：

#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>

int main() {
    int limit = 2000000;
    bool *numbers = (bool *)malloc(sizeof(bool) * limit);
    if (numbers == NULL) {
        printf("not enough memory\n");
        return 1;
    }
    for (int run = 0; run < limit; run++) {
        numbers[run] = true;
    }
    for (long long i = 2; i * i < limit; i++) {
        if (numbers[i] == true) {
            for (long long x = i * i; x < limit; x += i) {
                numbers[x] = false;
            }
        }
    }
    long long sum = 0;
    for (int run = 2; run < limit; run++) {
        if (numbers[run] == true) {
            sum = sum + run;
        }
    }
    printf("%lld\n", sum);
    free(numbers);
    return 0;
}

Answer 2

您可以尝试以下方式：

#include <stdio.h>
#include <stdlib.h>

int main(){
    unsigned long int i,j;  
    unsigned long long sum=0;         // the sum is potentially large
    unsigned long cnt=0;
    int limit=2000000;           // the limit for the sums
    char *primes;                // only needs to be used as a flag holder

    primes = malloc(sizeof(char)*limit);   // flag of is prime or not

    if(primes==NULL) {
        perror("main, malloc");
        return(-1);
     }    // else

    for (i=2;i<limit;i++)        // set the flags to True to start
        primes[i]=1;

    for (i=2;i<limit;i++)       // now go through all combos 
        if (primes[i])          // already know; no need
            for (j=i;i*j<limit;j++) // has i and j as factors 
                primes[i*j]=0;   // not prime

    for (i=2;i<limit;i++)      // now add them up
        if (primes[i]) {
            sum+=i;
            cnt++;
        }    
    // report what was found; note %lu for long or %llu for long long    
    printf("There are %lu primes less than %d with a sum of %llu",
             cnt, limit, sum);       

return 0;
}

您还可以使用 calloc 与 malloc 反转搜索以标记合数而不是质数并保存循环（感谢 Anti Haapala）：

#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>

int main(){
    unsigned long i,j;
    unsigned long long sum=0;        // the sum is potentially large
    unsigned long cnt=0;
    int limit=2000000;              // the limit for the sums

    bool * composite = (bool *)calloc(limit, sizeof(bool));  
    
    if(composite==NULL) {
        perror("main, calloc");
        return(-1);
    }

    for (i=2;i<limit;i++)       // now go through all combos 
        if (!composite[i])
            for (j=i;i*j<limit;j++) // has i and j as factors 
                composite[i*j]=true;   // composite; not prime

    for (i=2;i<limit;i++)      // now add them up
        if (!composite[i]) {
            sum+=i;
            cnt++;
        }    
    // report what was found; note %lu for long     
    printf("There are %lu primes less than %d with a sum of %llu", 
         cnt, limit, sum);       

return 0;
}

两者都打印：

There are 148933 primes less than 2000000 with a sum of 142913828922

注意：

对保证为 64 位的和使用 long long。长的可能是 32 位或 64 位，而 int 可能短至 16 位。 See C data types
根据要打印的数据类型，确保 printf type 和 length 说明符正确无误。

欧拉 10，素数之和到 2000000

2 个答案: