如何使用 for 循环制作一个每 5 个计数加起来为 5 的系列？

Question

我是 C 编程的新手，我正在尝试编写一个程序来打印系列的第 n 个项，每 5 个计数，它加起来为 5。

示例：1、2、3、4、9、10、11、12、17、18、19、20 ,25......

这是我的代码

int num,digit,count = 1;

printf("Enter n: ");
scanf("%d", &num);

for(int i=1; i<=num; i++){
    count++;
    if(count > 5){
        count = 0;
        i+=4;
    }
    
    printf("%d ",i);

}

我的代码没有达到我要求的特定第 n 个术语。例如，我输入了 10，它只显示到第 6 个学期

Answer 1

用这个替换你的循环，它也能工作：

for(int i=0, j=1, count=1; i<num; i++, j++, count++)
{
    printf("%d ", j);
    if(count == 4)
        count = 0, j+=4;
}

Answer 2

要做的事情就是在头脑中清楚自己想做什么以及如何去做。与其修改代码，不如将事情分解成简单的部分并使它们清晰。

#include <stdio.h>


int main(void)
{ 
    int num = 10;

    //  Method 1:  Spell everything out.

    //  i counts number of terms printed.
    //  j counts number of terms since last multiple of four terms.
    //  k is current term.
    for (
        int i = 0, j = 0, k = 1;    //  Initialize all counters.
        i < num;                    //  Continue until num terms are printed.
        ++i)                        //  Update count.
    {
        printf("%d ", k);           //  Print current term.
        ++j;                        //  Increment four-beat count.
        if (4 <= j)
        {
            //  Every fourth term, reset j and increment the term by 5.
            j = 0;
            k += 5;
        }
        else
            //  Otherwise, increment the term by 1.
            k += 1;
    }
    printf("\n");

    //  Method 2:  Use one counter and calculate term.

    //  Iterate i to print num terms.
    for (int i = 0; i < num; ++i)
        /*  Break the loop count into two parts:  the number of groups of 4
            (i/4) and a subcount within each group (i%4).  Looking at the
            starts of each group (1, 9, 17, 25...), we see each is eight
            greater than the previous one.  So we multiply the group number by
            8 to get the right offsets for them.  Within each group, the term
            increments by 1, so we use i%4 directly (effectively multiplied by
            1).  Then (i/4)*8 + i%4 would start us at 0 for i=0, but we want to
            start at 1, so we add 1.
        */
        printf("%d ", (i/4)*8 + i%4 + 1);
    printf("\n");
}

Answer 3

这是另一个我认为不那么复杂的观点，在评论中有解释：

#include <stdio.h>

int main(void)
{
    int num, count = 1;

    num = 20;
    // if you look closely, you actually have an initial condition before the
    // main pattern starts. Once the pattern starts, its actually every _fourth_
    // term that gets added by 5.  You'll make things easier on yourself if you
    // print out this initial condition, then handle the pattern in the loop.
    // If you really want to be correct, you can wrap this in a if (num > 0) check
    printf("%d ", count++);
    // start at 1, because we already printed the first item
    for(int i=1; i<num; i++, count++)
    {
        // now we can focus on every fourth term
        if (i % 4 == 0)
        {
            // if it's the fourth one, add 5 to the previous value
            // Of course this simplifies to count += 4
            count = (count-1) + 5;
        }
        
        printf("%d ",count);
    }
}

Demonstration

Answer 4

您不得更改 for 循环体内的变量 i。

您需要再引入一个变量来存储当前输出的数字。

这里有一个演示程序。

#include <stdio.h>

int main(void) 
{
    unsigned int n = 0;
    
    printf( "Enter n: " );
    
    scanf( "%u", &n );
    
    for ( unsigned int i = 0, value = 0, count = 1; i < n; i++ )
    {
        if ( count == 5 )
        {
            value += 5;
            count = 1;
        }
        else
        {
            ++value;
        }
        
        printf( "%u ", value );
        
        ++count;
    }
}

程序输出为

Enter n: 13
1 2 3 4 9 10 11 12 17 18 19 20 25

如何使用 for 循环制作一个每 5 个计数加起来为 5 的系列？

How to make a series that adds up by 5 every 5 counts using for loop?

c

for-loop