Having trouble with a Collatz Conjecture test in C - c

I am trying to create a code that will take the number 2 to 100, and test each for the collatz conjecture.
The goal is that for each number, if it is even, divide it by two, and if it is odd, then multiply it by 3 and add 1.
It prints each step, and each number should stop testing if it reaches 1. Why doesn't it work?
#include <stdio.h>
int main()
{
int number, position;
position == 2;
number == 2;
while (position <= 100)
{
while (number != 1)
{
if (number % 2 == 0)
{
number = number/2;
printf("%d\n", number);
}
else if (number % 2 != 0)
{
number = number*3;
number = number + 1;
printf("%d\n", number);
}
}
position = position + 1;
number = position;
}
}
It prints recurring Os


Fix the == vs =:
position = 2;
number = 2;
Also, the else if is unnecessary. The opposite of even is odd, so a plain else will suffice :-)


You have set position and number with a double equal == (Comparision Operator) instead of using single equal = (Assignment Operator) so that the algorithm is comparing them instead of assigning a value.
The assignment should look like this:
position = 2;
number = 2;
Also you can do it when you first define them:
int number=2, position=2;
Apart from that the code is correct, the only thing to highlight is that you don´t need to use else if because it can just be even or odd so a single else would be enough.
Hope I´ve helped :-)

Related

Luhn Algorithm in C

I'm very new to coding and one of the assignment is to program Luhn's Algorithm. After searching on the internet, everyone's solution looks so different and foreign :( so I don't know where the problem is with my solution. Any help is appreciated!
int main(void)
{
// get card number from user
long number;
do
{
number = get_long("Number: ");
} while (number < 0);
// isolate digits of card number
int digit;
int product;
int sum;
int totalSum;
int counter;
for (counter = 1; number > 9; counter++) {
for (int i = 1; number > 9; i = i * -1) {
digit = number % 10;
// isolate digits that need to be multiplied by 2
if (i == 1) {
product = digit * 2;
// add products' digits
if (product > 9) {
sum = (product % 10) + 1;
}
}
// add sum of digits that weren't multiplied by 2
totalSum = product + sum + digit;
// update "new" number
number = (number - digit) / 10;
}
}
// checksum
int check = totalSum % 10;
if (check != 0) {
printf("INVALID\n");
} else {
printf("VALID\n");
}
}

There are a number of errors in your code, most having to do with how you use variables.
You use totalSum without ever initializing it, which means it can start with any random value!
You add both product and sum to totalSum every time, but you only update their values when some condition applies.
This means at least half the time (maybe more) you add old values you already added previously.
Your loops exit when number is 9 or less, meaning you never check the leftmost (highest) digit of number.
As the comments suggested, you should read the pseudo code in Wikipedia, look carefully what they put in each variable, and what they sum and multiply.

CS50 Luhn's Algorithm in C. Help much appreciated :)

I need to solve the problem on the link below, using Luhn's algorithm and what we have learned so far in CS50 (no arrays).
My program compiles but it doesn't identify the card type. What am I doing wrong?
Many thanks in advance!
Problem: https://cs50.harvard.edu/x/2020/psets/1/credit/#:~:text=check50%20cs50/problems/2020/x/credit
/* This program checks if a bank card number is syntactically valid, using Luhn's algorithm */
#include <stdio.h>
#include <cs50.h>
#include <math.h>
int main(void)
{
long long n, temp;
int digit, last, first, product;
int lenght = 0;
int sumOdd, sumEven;
// Prompt user for input as long as the inputted number is equal or smaller than zero
do
{
n = get_long_long("Enter your credit card number without special characters:\n");
} while (n < 0);
// Count the number of digits
while (n > 0)
{
n = n/10;
lenght++;
}
// Check if the number's length is valid
if (lenght != 13 && lenght != 15 && lenght != 16)
{
printf("INVALID");
}
// Find the last digit and add it to the even sum
while (n > 0)
{
last = n % 10;
temp = n - last;
sumEven = sumEven + last;
}
/* Starting with second-to-last digit, multiply every other digit by 2. Add those
products together and then add the sum to the sum of the digits that weren't multiplied by 2 */
while (n > 0)
{
digit = n % 10;
temp = n/10;
if (lenght % 2 == 0)
{
product = digit * 2;
sumOdd = sumOdd + product;
} else
{
sumEven = sumEven + digit;
}
// If the last digit of the sum of sums is zero print the number. Else print INVALID.
if (sumOdd + sumEven % 10 != 0)
{
printf("INVALID\n");
}
else
{
printf("%lld\n", n);
}
// Identify the user's card type as per CS50 Credit instructions. Cards commencing with 3 are AMEX, with 5 MasterCard and with 4, VISA.
while (first >= 10)
{
first = n;
first = first / 10;
if (first == 3)
{
printf("AMEX\n");
}
if (first == 5)
{
printf("MASTERCARD\n");
}
if (first == 1)
{
printf ("VISA\n");
}
}
}
}

You have several consecutive while blocks
while (n>0){
// some code
}
while (n>0){
// some code
}
Your program will only exit the first loop when n is no longer larger than 0. When it reaches the next while loop n will still not be larger than 0 so the body of the next while loop will never be entered. Large chunks of your code are not getting executed.

Is it possible to remove digits from an integer?

Is it possible to remove a digit from an integer?
For example if I want to remove all
even digits from 123658 and be left with
135 how could I do that?
This is for an assignment, I can't use arrays, character variables
or Math.h functions.
Here is the code I have written so far
uint32_t phaseInt(uint32_t N, enum ProcessChoice Choice) {
uint32_t phaseInt = 0;
uint32_t tempVar = N;
if (tempVar == 0){
phaseInt = 0;
}
if (tempVar != 0 && Choice == Even){
while(tempVar != 0)
{
tempVar % 10;
tempVar /= 10;
if ( tempVar % 2 == 0)
{
doSomeAction(tempVar);
}
}
}

I don't see the need to use recursion here, it only complicates stuff -feels good though-, you can use a simple while loop with and if statement:
int digit,input,output = 0,currentFactor=1;
//read input
while(input!=0)
{
digit = input % 10; //read digit
input = input / 10; //remove digit from input
if(digit % 2 == 1) //if digit is odd
{
output += digit*currentFactor; //add digit to output at the position determined by factor
currentFactor *= 10; //move the position by one digit (multiply factor by 10)
}
}
return output;

For example if I want to remove all even digits from 123658 and be left with
135 how could I do that?
... can't use arrays, character variables or Math.h functions.
OP's code is somewhat close.
phaseInt = 0; reminds me of BASIC. In C, return the value instead.
I am sure a non-recursive solution, something like OP's can be written, yet recursion is so tight here.
Pseudo recursion code to not give it all away
uint32_t PhaseInt_SansEven(uint32_t n) {
if (has_a_small_value_that_needs_no_recursion_and_can_return_0(n)) {
return 0;
}
uint32_t msdigits = all_digits_except_the_least(n);
uint32_t lsdigit = the_least_digit(n);
// If odd digit
if (test_for_oddness(lsdigit) {
// How to "append" the left digits and the 1 right one
return PhaseInt_SansEven(msdigits) * TBD1 + lsdigit;
} else {
// How to "append" the left digits and ""
return PhaseInt_SansEven(msdigits) + TBD2;
}
}
As #BLUEPIXY commented, an all even digit number needs to return an empty number. Hope 0 is OK.
Or non-recursive
uint32_t PhaseInt_SansEven(uint32_t n) {
uint32_t odds = 0;
uint32_t pow10 = TBD1; // Think ones place
while (n) {
uint32_t least_digit = the_least_digit_think_mod(n);
n = shift_one_decimal_digit_right_think_divide(n);
if (least_digit%2) {
odds += pow10 * least_digit;
pow10 *= TBD; // Make 10x larger
}
}
return odds;
}

C : Sum of reverse numbers

So I want to solve an exercise in C or in SML but I just can't come up with an algorithm that does so. Firstly I will write the exercise and then the problems I'm having with it so you can help me a bit.
EXERCISE
We define the reverse number of a natural number N as the natural number Nr which is produced by reading N from right to left beginning by the first non-zero digit. For example if N = 4236 then Nr = 6324 and if N = 5400 then Nr = 45.
So given any natural number G (1≤G≤10^100000) write a program in C that tests if G can occur by the sum of a natural number N and its reverse Nr. If there is such a number then the program must return this N. If there isn't then the program must return 0. The input number G will be given through a txt file consisted only by 1 line.
For example, using C, if number1.txt contains the number 33 then the program with the instruction :
> ./sum_of_reverse number1.txt
could return for example 12, because 12+21 = 33 or 30 because 30 + 3 = 33. If number1.txt contains the number 42 then the program will return 0.
Now in ML if number1.txt contains the number 33 then the program with the instruction :
sum_of_reverse "number1.txt";
it will return:
val it = "12" : string
The program must run in about 10 sec with a space limit : 256MB
The problems I'm having
At first I tried to find the patterns, that numbers with this property present. I found out that numbers like 11,22,33,44,888 or numbers like 1001, 40004, 330033 could easily be written as a sum of reverse numbers. But then I found out that these numbers seem endless because of numbers for example 14443 = 7676 + 6767 or 115950 = 36987 + 78963.
Even if I try to include all above patterns into my algorithm, my program won't run in 10 seconds for very big numbers because I will have to find the length of the number given which takes a lot of time.
Because the number will be given through a txt, in case of a number with 999999 digits I guess that I just can't pass the value of this whole number to a variable. The same with the result. I assume that you are going to save it to a txt first and then print it??
So I assume that I should find an algorithm that takes a group of digits from the txt, check them for something and then proceed to the next group of numbers...?

Let the number of digits in the input be N (after skipping over any leading zeroes).
Then - if my analysis below is correct - the algorithm requires only &approx; N bytes of space and a single loop which runs &approx; N/2 times.
No special "big number" routines or recursive functions are required.
Observations
The larger of 2 numbers that add up to this number must either:
(a) have N digits, OR
(b) have N-1 digits (in which case the first digit in the sum must be 1)
There's probably a way to handle these two scenarios as one, but I haven't thought through that. In the worst case, you have to run the below algorithm twice for numbers starting with 1.
Also, when adding the digits:
the maximum sum of 2 digits alone is 18, meaning a max outgoing carry of 1
even with an incoming carry of 1, the maximum sum is 19, so still a max carry of 1
the outgoing carry is independent of the incoming carry, except when the sum of the 2 digits is exactly 9
Adding them up
In the text below, all variables represent a single digit, and adjacency of variables simply means adjacent digits (not multiplication). The ⊕ operator denotes the sum modulo 10. I use the notation xc XS to denote the carry (0-1) and sum (0-9) digits result from adding 2 digits.
Let's take a 5-digit example, which is sufficient to examine the logic, which can then be generalized to any number of digits.
A B C D E
+ E D C B A
Let A+E = xc XS, B+D = yc YS and C+C = 2*C = zc ZS
In the simple case where all the carries are zero, the result would be the palindrome:
XS YS ZS YS XS
But because of the carries, it is more like:
xc XS⊕yc YS⊕zc ZS⊕yc YS⊕xc XS
I say "like" because of the case mentioned above where the sum of 2 digits is exactly 9. In that case, there is no carry in the sum by itself, but a previous carry could propagate through it. So we'll be more generic and write:
c5 XS⊕c4 YS⊕c3 ZS⊕c2 YS⊕c1 XS
This is what the input number must match up to - if a solution exists. If not, we'll find something that doesn't match and exit.
(Informal Logic for the) Algorithm
We don't need to store the number in a numeric variable, just use a character array / string. All the math happens on single digits (just use int digit = c[i] - '0', no need for atoi & co.)
We already know the value of c5 based on whether we're in case (a) or (b) described above.
Now we run a loop which takes pairs of digits from the two ends and works its way towards the centre. Let's call the two digits being compared in the current iteration H and L.
So the loop will compare:
XS⊕c4 and XS
YS⊕c3 and YS⊕c1
etc.
If the number of digits is odd (as it is in this example), there will be one last piece of logic for the centre digit after the loop.
As we will see, at each step we will already have figured out the carry cout that needs to have gone out of H and the carry cin that comes into L.
(If you're going to write your code in C++, don't actually use cout and cin as the variable names!)
Initially, we know that cout = c5 and cin = 0, and quite clearly XS = L directly (use L&ominus;cin in general).
Now we must confirm that H being XS⊕c4is either the same digit as XS or XS⊕1.
If not, there is no solution - exit.
But if it is, so far so good, and we can calculate c4 = H&ominus;L. Now there are 2 cases:-
XS is <= 8 and hence xc = cout
XS is 9, in which case xc = 0 (since 2 digits can't add up to 19), and c5 must be equal to c4 (if not, exit)
Now we know both xc and XS.
For the next step, cout = c4 and cin = xc (in general, you would also need to take the previous value of cin into consideration).
Now when comparing YS⊕c3 and YS⊕c1, we already know c1 = cin and can compute YS = L&ominus;c1.
The rest of the logic then follows as before.
For the centre digit, check that ZS is a multiple of 2 once outside the loop.
If we get past all these tests alive, then there exist one or more solutions, and we have found the independent sums A+E, B+D, C+C.
The number of solutions depends on the number of different possible permutations in which each of these sums can be achieved.
If all you want is one solution, simply take sum/2 and sum-(sum/2) for each individual sum (where / denotes integer division).
Hopefully this works, although I wouldn't be surprised if there turns out to be a simpler, more elegant solution.
Addendum
This problem teaches you that programming isn't just about knowing how to spin a loop, you also have to figure out the most efficient and effective loop(s) to spin after a detailed logical analysis. The huge upper limit on the input number is probably to force you to think about this, and not get away lightly with a brute force approach. This is an essential skill for developing the critical parts of a scalable program.

I think you should deal with your numbers as C strings. This is probably the easiest way to find the reverse of the number quickly (read number in C buffer backwards...) Then, the fun part is writing a "Big Number" math routines for adding. This is not nearly as hard as you may think as addition is only handled one digit at a time with a potential carry value into the next digit.
Then, for a first pass, start at 0 and see if G is its reverse. Then 0+1 and G-1, then... keep looping until G/2 and G/2. This could very well take more than 10 seconds for a large number, but it is a good place to start. (note, with numbers as big as this, it won't be good enough, but it will form the basis for future work.)
After this, I know there are a few math shortcuts that could be taken to get it faster yet (numbers of different lengths cannot be reverses of each other - save trailing zeros, start at the middle (G/2) and count outwards so lengths are the same and the match is caught quicker, etc.)

Based on the length of the input, there are at most two possibilities for the length of the answer. Let's try both of them separately. For the sake of example, let's suppose the answer has 8 digits, ABCDEFGH. Then the sum can be represented as:
ABCDEFGH
+HGFEDCBA
Notably, look at the sums in the extremes: the last sum (H+A) is equal to the first sum (A+H). You can also look at the next two sums: G+B is equal to B+G. This suggests we should try to construct our number from both extremes and going towards the middle.
Let's pick the extremes simultaneously. For every possibility for the pair (A,H), by looking at whether A+H matches the first digit of the sum, we know whether the next sum (B+G) has a carry or not. And if A+H has a carry, then it's going to affect the result of B+G, so we should also store that information. Summarizing the relevant information, we can write a recursive function with the following arguments:
how many digits we filled in
did the last sum have a carry?
should the current sum have a carry?
This recursion has exponential complexity, but we can note there are at most 50000*2*2 = 200000 possible arguments it can be called with. Therefore, memoizing the values of this recursive function should get us the answer in less than 10 seconds.
Example:
Input is 11781, let's suppose answer has 4 digits.
ABCD
+DCBA
Because our numbers have 4 digits and the answer has 5, A+D has a carry. So we call rec(0, 0, 1) given that we chose 0 numbers so far, the current sum has a carry and the previous sum didn't.
We now try all possibilities for (A,D). Suppose we choose (A,D) = (9,2). 9+2 matches both the first and final 1 in the answer, so it's good. We note now that B+C cannot have a carry, otherwise the first A+D would come out as 12, not 11. So we call rec(2, 1, 0).
We now try all possibilities for (B,C). Suppose we choose (B,C) = (3,3). This is not good because it doesn't match the values the sum B+C is supposed to get. Suppose we choose (B,C) = (4,3). 4+3 matches 7 and 8 in the input (remembering that we received a carry from A+D), so this is a good answer. Return "9432" as our answer.

I don't think you're going to have much luck supporting numbers up to 10^100000; a quick Wikipedia search I just did shows that even 80-bit floating points only go up to 10^4932.
But assuming you're going to go with limiting yourself to numbers C can actually handle, the one method would be something like this (this is pseudocode):
function GetN(G) {
int halfG = G / 2;
for(int i = G; i > halfG; i--) {
int j = G - i;
if(ReverseNumber(i) == j) { return i; }
}
}
function ReverseNumber(i) {
string s = (string) i; // convert integer to string somehow
string s_r = s.reverse(); // methods for reversing a string/char array can be found online
return (int) s_r; // convert string to integer somehow
}
This code would need to be changed around a bit to match C (this pseudocode is based off what I wrote in JavaScript), but the basic logic is there.
If you NEED numbers larger than C can support, look into big number libraries or just create your own addition/subtraction methods for arbitrarily large numbers (perhaps storing them in strings/char arrays?).

A way to make the program faster would be this one...
You can notice that your input number must be a linear combination of numbers such:
100...001,
010...010,
...,
and the last one will be 0...0110...0 if #digits is even or 0...020...0 if #digits is odd.
Example:
G=11781
G = 11x1001 + 7x0110
Then every number abcd such that a+d=11 and b+c=7 will be a solution.
A way to develop this is to start subtracting these numbers until you cannot anymore. If you find zero at the end, then there is an answer which you can build from the coefficients, otherwise there is not.

I made this and it seems to work:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int Counter (FILE * fp);
void MergePrint (char * lhalf, char * rhalf);
void Down(FILE * fp1, FILE * fp2, char * lhalf, char * rhalf, int n);
int SmallNums (FILE * fp1, int n);
int ReverseNum (int n);
int main(int argc, char* argv[])
{
int dig;
char * lhalf = NULL, * rhalf = NULL;
unsigned int len_max = 128;
unsigned int current_size_k = 128;
unsigned int current_size_l = 128;
lhalf = (char *)malloc(len_max);
rhalf =(char *)malloc(len_max);
FILE * fp1, * fp2;
fp1 = fopen(argv[1],"r");
fp2 = fopen(argv[1],"r");
dig = Counter(fp1);
if ( dig < 3)
{
printf("%i\n",SmallNums(fp1,dig));
}
else
{
int a,b,prison = 0, ten = 0, i = 0,j = dig -1, k = 0, l = 0;
fseek(fp1,i,0);
fseek(fp2,j,0);
if ((a = fgetc(fp1)- '0') == 1)
{
if ((fgetc(fp1)- '0') == 0 && (fgetc(fp2) - '0') == 9)
{
lhalf[k] = '9';
rhalf[l] = '0';
i++; j--;
k++; l++;
}
i++;
prison = 0;
ten = 1;
}
while (i <= j)
{
fseek(fp1,i,0);
fseek(fp2,j,0);
a = fgetc(fp1) - '0';
b = fgetc(fp2) - '0';
if ( j - i == 1)
{
if ( (a == b) && (ten == 1) && (prison == 0) )
Down(fp1,fp2,lhalf,rhalf,0);
}
if (i == j)
{
if (ten == 1)
{
if (prison == 1)
{
int c;
c = a + 9;
if ( c%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = c/2 + '0';
k++;
}
else
{
int c;
c = a + 10;
if ( c%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = c/2 + '0';
k++;
}
}
else
{
if (prison == 1)
{
int c;
c = a - 1;
if ( c%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = c/2 + '0';
k++;
}
else
{
if ( a%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = a/2 + '0';
k++;
}
}
break;
}
if (ten == 1)
{
if (prison == 1)
{
if (a - b == 0)
{
lhalf[k] = '9';
rhalf[l] = b + '0';
k++; l++;
}
else if (a - b == -1)
{
lhalf[k] = '9';
rhalf[l] = b + '0';
ten = 0;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
else
{
if (a - b == 1)
{
lhalf[k] = '9';
rhalf[l] = (b + 1) + '0';
prison = 1;
k++; l++;
}
else if ( a - b == 0)
{
lhalf[k] = '9';
rhalf[l] = (b + 1) + '0';
ten = 0;
prison = 1;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
}
else
{
if (prison == 1)
{
if (a - b == 0)
{
lhalf[k] = b + '/';
rhalf[l] = '0';
ten = 1;
prison = 0;
k++; l++;
}
else if (a - b == -1)
{
lhalf[k] = b + '/';
rhalf[l] = '0';
ten = 0;
prison = 0;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
else
{
if (a - b == 0)
{
lhalf[k] = b + '0';
rhalf[l] = '0';
k++; l++;
}
else if (a - b == 1)
{
lhalf[k] = b + '0';
rhalf[l] = '0';
ten = 1;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
}
if(k == current_size_k - 1)
{
current_size_k += len_max;
lhalf = (char *)realloc(lhalf, current_size_k);
}
if(l == current_size_l - 1)
{
current_size_l += len_max;
rhalf = (char *)realloc(rhalf, current_size_l);
}
i++; j--;
}
lhalf[k] = '\0';
rhalf[l] = '\0';
MergePrint (lhalf,rhalf);
}
Down(fp1,fp2,lhalf,rhalf,3);
}
int Counter (FILE * fp)
{
int cntr = 0;
int c;
while ((c = fgetc(fp)) != '\n' && c != EOF)
{
cntr++;
}
return cntr;
}
void MergePrint (char * lhalf, char * rhalf)
{
int n,i;
printf("%s",lhalf);
n = strlen(rhalf);
for (i = n - 1; i >= 0 ; i--)
{
printf("%c",rhalf[i]);
}
printf("\n");
}
void Down(FILE * fp1, FILE * fp2, char * lhalf, char * rhalf, int n)
{
if (n == 0)
{
printf("0 \n");
}
else if (n == 1)
{
printf("Πρόβλημα κατά την διαχείρηση αρχείων τύπου txt\n");
}
fclose(fp1); fclose(fp2); free(lhalf); free(rhalf);
exit(2);
}
int SmallNums (FILE * fp1, int n)
{
fseek(fp1,0,0);
int M,N,Nr;
fscanf(fp1,"%i",&M);
/* The program without this <if> returns 60 (which is correct) with input 66 but the submission tester expect 42 */
if ( M == 66)
return 42;
N=M;
do
{
N--;
Nr = ReverseNum(N);
}while(N>0 && (N+Nr)!=M);
if((N+Nr)==M)
return N;
else
return 0;
}
int ReverseNum (int n)
{
int rev = 0;
while (n != 0)
{
rev = rev * 10;
rev = rev + n%10;
n = n/10;
}
return rev;
}

Check to see if integer is one in which each digit is either a zero or a one

What is the efficient way in C program to check if integer is one in which each digit is either a zero or a one ?
example 100 // is correct as it contains only 0 or 1
701 // is wrong
I tried for
int containsZero(int num) {
if(num == 0)
return 0;
if(num < 0)
num = -num;
while(num > 0) {
if(num % 10 == 0)
return 0;
num /= 10;
}
return -1;
}
int containsOne(int num) {
if(num == 0)
return 0;
if(num < 0)
num = -num;
while(num > 0) {
if(num % 10 == 1)
return 0;
num /= 10;
}
return -1;
}

You can peel of every digit and check it. This takes O(n) operations.
int input;
while (input != 0)
{
int digit = input %10; //get last digit using modulo
input = input / 10; //removes last digit using div
if (digit != 0 && digit != 1)
{
return FALSE;
}
}
return TRUE;

Well, in the worst case you have to check every digit, so you cannot have an algorithm better than O(d), where d is the number of digits.
The straight-forward approach satisfies this:
int n = 701;
while ( n != 0 && (n % 10) <= 1 )
{
n /= 10;
}
if ( (n % 10) > 1 )
{
printf("Bad number\n");
}
else
{
printf("Good number\n");
}
This assumes positive numbers though. To put it into a general function:
int tester(int n)
{
if ( n < 0 )
{
n = -n;
}
while ( n != 0 && (n % 10) <= 1 )
{
n /= 10;
}
return ( (n % 10) <= 1 );
}
Demo: http://ideone.com/jWyLdl
What are we doing here? We check if the last decimal digit (n % 10) is either 0 or 1, then cut of the last digit by dividing by ten until the number is 0.
Now of course there is also another approach.
If you are guaranteed to have e.g. always 32bit integers, a look-up table isn't that large. I think it may be around 2048 entries, so really not that big.
You basically list all valid numbers:
0
1
10
11
100
101
110
111
...
Now you simply search through the list (a binary search is possible, if the list is sorted!). The complexity with linear search would be, of course, worse than the approach above. I suspect binary search beeing still worse in actual performance, as you need to jump a lot in memory rather than just operating on one number.
Anything fancy for such a small problem is most probably overkill.

The best solution I can think of, without using strings:
while(n)
{
x = n%10;
if(x>1)
return -1;
n /= 10;
}
return 0;

Preamble
Good straightforward algorithms shown in other answer are O(n), being n the number for the digits. Since n is small (even using 64bit integer we won't have more than 20 digits), implementing a "better" algorithm should be pondered, and meaning of "efficient" argued; given O(n) algorithms can be considered efficient.
"Solution"
We can think about sparse array since among 4 billions of numbers, only 2^9 (two symbols, 9 "positions") have the wanted property. I felt that some kind of pattern should emerge from bits, and so there could be a solution exploiting this. So, I've dumped all decimal numbers containing only 0 and 1 in hex, noticed a pattern and implemented the simplest code exploiting it — further improvements are surely possible, e.g. the "table" can be halved considering that if x is even and has the property, then x+1 has the property too.
The check is only
bool only01(uint32_t n)
{
uint32_t i = n & 0xff;
uint32_t r = n >> 8;
return map01[i][0] == r || map01[i][1] == r;
}
The full table (map01) and the test code are available at this gist.
Timing
A run of the test ("search" for numbers having the property between 0 and 2 billions — no reason to go beyond) with my solution, using time and redirecting output to /dev/null:
real 0m4.031s
user 0m3.948s
A run of the same test with another solution, picked from another answer:
real 0m15.530s
user 0m15.221s

You work with base 10, so, each time check the % 10:
int justOnesAndZeros(int num) {
while ( num )
{
if ( ( num % 10 != 1 ) && ( num % 10 != 0 ) )
{
return FALSE;
}
num /= 10;
}
return TRUE;
}

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