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C Program To Accept The String That Ends With 011 | C Programming

C Program To Accept The String That Ends With 011

#include <stdio.h>
#include <string.h>
int main()
{
    char str[10];
    int len, i, q = 0;
    printf("Enter the string which only contain 0 and 1\n");
    scanf("%s", str);
    len = strlen(str);
    for (i = 0; i <= len; i++)
    {
        if (str[i] == '0' && q == 0)
            q = 1;
        else if (str[i] == '1' && q == 0)
            q = 0;
        else if (str[i] == '1' && q == 1)
            q = 2;
        else if (str[i] == '0' && q == 1)
            q = 1;
        else if (str[i] == '0' && q == 2)
            q = 1;
        else if (str[i] == '1' && q == 2)
            q = 3;
        else if (str[i] == '1' && q == 3)
            q = 0;
        else if (str[i] == '0' && q == 3)
            q = 1;
    }
    if (q == 3)
        printf("Given string is accepted\n");
    else
        printf("Given string is not accepted\n");
}

Some Of The Outputs Of Above Program

OUTPUT #1

C Program To Accept The String That Ends With 011

OUTPUT #2

C Program To Accept The String That Ends With 011

OUTPUT #3

C Program To Accept The String That Ends With 011

OUTPUT #4

C Program To Accept The String That Ends With 011

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