c - 在单独的行上打印 10 范围内的素数

Question

我想为每 10 个输入分离程序的输出(打印)。代码是求1到n中的素数以及它们的中位数。

我只想将 n 的每 10 个计数分开，例如:

primes from  1 to 10:
       from 10 to 20:
       from 20 to 30:

你能帮我做一下吗？

#include <stdio.h>

int main()
{
    int n, x, y, p, sum, sumx;
    float arm;
    sum = 0, sumx = 0;
    printf("\n Enter the maximum value of n =");
    scanf("%d", &n);
    printf("\n All prime numbers from 1 to %d are: ", n);

    for (x = 2; x <= n; x++)
    {
        p = 1;
        for (y = 2; y <= x / 2; y++)
        {
            if (x % y == 0)
            {
                p = 0;
                break;
            }
        }
        if (p == 1)
        {
            sum += x;
        }
        if (p == 1 && x != 1)
        {
            printf("%d ", x);
            sumx++;
            arm = (sum * 1.0) / sumx;
        }
    }

    printf("\nThe sum of all Prime numbers from 1 to %d =%d", n, sum);
    printf("\nTotal prime numbers is %d", sumx);
    printf("\nThe median of all Prime numbers from 1 to %d is %.2f", n, arm);

    return 0;
}

Answer 1

只要 x 的值能被 10 整除，您就可以打印出“From _ to _:”:

#include <stdio.h>

int main()
{
    int n, x, y, p, sum, sumx;
    float arm;
    sum = 0, sumx = 0;
    printf("\n Enter the maximum value of n =");
    scanf("%d", &n);
    printf("\n All prime numbers from 1 to %d are: ", n);
    int upper = n < 10? n : 10;
    printf("\nFrom 1 to %d: ", upper);

    for (x = 2; x <= n; x++)
    {
        p = 1;
        for (y = 2; y <= x / 2; y++)
        {
            if (x % y == 0)
            {
                p = 0;
                break;
            }
        }
        if (p == 1)
        {
            sum += x;
        }
        if (p == 1 && x != 1)
        {
            printf("%d ", x);
            sumx++;
            arm = (sum * 1.0) / sumx;
        }

        if (x % 10 == 0 && x != n) {
            int upper = (x + 10) > n? n : (x + 10);
            printf("\nFrom %d to %d: ", x + 1, upper);
        }   
    }

    printf("\nThe sum of all Prime numbers from 1 to %d =%d", n, sum);
    printf("\nTotal prime numbers is %d", sumx);
    printf("\nThe median of all Prime numbers from 1 to %d is %.2f", n, arm);

    return 0;
}

这将导致如下输出:

Enter the maximum value of n =29

All prime numbers from 1 to 29 are:

From 1 to 10: 2 3 5 7

From 11 to 20: 11 13 17 19

From 21 to 29: 23 29

The sum of all Prime numbers from 1 to 29 =129

Total prime numbers is 10

The median of all Prime numbers from 1 to 29 is 12.90