将第一句中所有单词的寄存器更改为第二句中单词的寄存器

Question

任务是更改第一句中所有与第二句中的单词相似的单词的寄存器。寄存器应与第二句中的相同。

这是我到目前为止获得的代码，但是我有点卡住了。 也许有更简单的方法可以不用数组来做到这一点。谢谢。

int main()
{
    char input[100];
    char input2[100];
    char cheking[100];
    char cheking2[100]
    int i, j, j1, j2, l; //COUNTERS
    int k;

    printf("Please, enter the 1st sentence:\n");
    for (i = 0; (input[i] = getchar()) != '\n'; i++)
    {
        ;
    }
    printf("Please, enter the 2nd sentence:\n");
    for (i = 0; (input2[i] = getchar()) != '\n'; i++)
    {
        ;
    }
    i = 0;
    j1 = strlen(input) - 1; //CHECK AFTER
    j2 = strlen(input2) - 1;

    //do
    for (i = 0; i < j1; i++) //I didn't realise it for every word of the 1st sentence
        for (k = 0; (input[k] != ' '); k++)
        {
            checking[k] = tolower(input[k]);
        }

    for (i = 0; i < j2; i++)
    {
        for (k = 0; (input2[k] != '\n'); k++)
        {
            checking2[k] = tolower(input2[k]);
        }
    }

    for(i = 0; i < j2; i++)
    {
        for (j = 0; (checking2[j] != ' '), j++)
        {
            if (cheking1[j] != cheking2[j])
            {
                i++;
                j = 0;
                k = 1;
            }
            else (k = 0);
        }

        if (k == 1)
        for(i; (input[i] != ' '); i++) //don't know how to change word to that it should be 
        {

        }
    }

    //while ((input[i] != '\n') || (input[2] != '\n');



    getch();
    return 0;
}