用c编程以打印flloyd的三角形。 程序停止工作

Question

i wrote the following code in c to print floyd's triangle.

int main()
{
    printf("Enter the number of rows you want to have");
    int t;
    scanf("%d",&t);
    int i;
    char a[1000] ="";
    for(i=1;i<=t;i++)
    {
        if (i%2!=0)
        {
            strcat("1",a);
            printf("%c\n",a);}
        else
            strcat("0",a);
        printf("%c\n",a);


    }
    return 0;
}

The program seems fine to me but it stops working as soon as i execute it. Please help I want to have the output as follows-
1
01
101
0101
10101
and so on

Answer 1

You can construct the string (the bigger one) first and then print only a part of it in each row:

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

int main()
{
    printf("Enter the number of rows you want to have");
    int t;
    scanf("%d",&t); // You should check the return value...
    puts("");

    char pattern[t + 1]; // It's a VLA, so you need a C99 compliant compiler or
                         // use your big array...

    // Initialize the string (it's the last row) starting from the last 
    // char (the null terminator) and stepping back to first. Every row should
    // end with a '1', so in the loop, start with it.
    int i = t;
    pattern[i] = '\0';
    while ( i > 0 )
    {
        pattern[--i] = '1';
        if ( i == 0 )
            break;
        pattern[--i] = '0';
    }

    // Print only the part needed in each row
    char* tmp = &pattern[t - 1];
    for ( int i = 0; i < t; ++i, --tmp )
    {
        printf("%s\n", tmp);
    }
    return 0;
}

Answer 2

Compile with warnings enabled and you will quickly see that you need to print a with %s (string format), rather than %c (character format). When I compiled your code, I got:

prog.c: In function 'main':
prog.c:16:22: warning: format '%c' expects argument of type 'int', but argument 2 has type 'char *' [-Wformat=]
             printf("%c\n",a);
                     ~^    ~
                     %s
prog.c:20:22: warning: format '%c' expects argument of type 'int', but argument 2 has type 'char *' [-Wformat=]
             printf("%c\n",a);
                     ~^    ~
                     %s

Moreover, your else statement lacks curly braces, which results in only the strcat() to be assumed as its body.

To get the desired output, you should abandon strcat() and index the position you want to assign the bit, like this:

#include <stdio.h>
#include <string.h>

int main()
{
    printf("Enter the number of rows you want to have\n");
    int t;
    scanf("%d",&t);
    int i;
    char a[1000] ="";
    for(i=1;i<=t;i++)
    {
        if (i%2!=0)
        {
            a[999 - i] = '1';
            printf("%s\n", &a[999 - i]);
        }
        else {
            a[999 - i] = '0';
            printf("%s\n", &a[999 - i]);
        }
    }
    return 0;
}

Output:

Enter the number of rows you want to have
4
1
01
101
0101

Notice that 999 is the size of your array, minus 1.

---

PS: In your posted code: when concatenating the string, you messed up the order of the arguments.

Answer 3

This should give you the output you're looking for:

#include <iostream>
#include <string.h>

int main()
{
    printf("Enter the number of rows you want to have: ");
    int t;
    scanf("%d", &t);

    for (int i = 1; i <= t; i++)
    {
        for (int j = 1; j <= i; j++)
        {
            char a[1000] = "";

            if ((i+j) % 2 == 0)
                strcat(a, "1");
            else
                strcat(a, "0");

            printf("%s", a);
        }

        printf("\n");
    }

    return 0;
}

Since every other line begins with a 0, you could simply recreate the string a per line.