I've created a solution to problem 4 on Project Euler.
However, what I find is that placing the print statement (that prints the answer) in different locations prints different answers. And for some reason, the highest value of result is 580085. Shouldn't it be 906609? Is there something wrong with my isPalindrome() method?
#include <stdio.h>
#include <stdbool.h>
int isPalindrome(int n);
//Find the largest palindrome made from the product of two 3-digit numbers.
int main(void)
{
int i = 0;
int j = 0;
int result = 0;
int palindrome = 0;
int max = 0;
//Each iteration of i will be multiplied from j:10-99
for(i = 100; i <= 999; i++)
{
for(j = 100; j <= 999; j++)
{
result = i * j;
if(isPalindrome(result) == 0)
{
//printf("Largest Palindrome: %d\n", max); //906609
//printf("Result: %d\n", result); //580085
if(result > max)
{
max = result;
//printf("Largest Palindrome: %d\n", max); //927340
}
printf("Largest Palindrome: %d\n", max); //906609
}
}
}
//printf("Largest Palindrome: %d\n", max); //998001
system("PAUSE");
return 0;
} //End of main
//Determines if number is a palindrome
int isPalindrome(int num)
{
int n = num;
int i = 0;
int j = 0;
int k = 0;
int count = 0;
int yes = 0;
//Determines the size of numArray
while(n/10 != 0)
{
n%10;
count++;
n = n/10;
}
int numArray[count];
//Fill numArray with each digit of num
for(i = 0; i <= count; i++)
{
numArray[i] = num%10;
//printf("%d\n", numArray[i]);
num = num/10;
}
//Determines if num is a Palindrome
while(numArray[k] == numArray[count])
{
k = k + 1;
count = count - 1;
yes++;
}
if(yes >= 3)
{
return 0;
}
}//End of Function
I remember doing that problem a while ago and I simply made a is_palindrome() function and brute-forced it. I started testing from 999*999 downwards.
My approach to detect a palindrome was rather different from yours. I would convert the given number to a string and compare the first char with the nth char, second with n-1 and so on.
It was quite simple (and might be inefficient too) but the answer would come up "instantly".
There is no problem in the code in finding the number.
According to your code fragment:
.
.
}
printf("Largest Palindrome: %d\n", max); //906609
}
}
}
//printf("Largest Palindrome: %d\n", max); //998001
system("PAUSE");
.
.
.
you get the result as soon as a palindrome number is found multiplying the number downwards.
You should store the palindrome in a variable max and let the code run further as there is a possibility of finding a greater palindrome further.
Note that for i=800,j=500, then i*j will be greater when compare with i=999,j=100.
Just understand the logic here.
A few issues in the isPalindrome function :
the first while loop doesn't count the number of digits in the number, but counts one less.
as a result, the numArray array is too small (assuming that your compiler supports creating the array like that to begin with)
the for loop is writing a value past the end of the array, at best overwriting some other (possibly important) memory location.
the second while loop has no properly defined end condition - it can happily compare values past the bounds of the array.
due to that, the value in yes is potentially incorrect, and so the result of the function is too.
you return nothing if the function does not detect a palindrome.
//Determines if number is a palindrome
bool isPalindrome(int num) // Change this to bool
{
int n = num;
int i = 0;
int j = 0;
int k = 0;
int count = 1; // Start counting at 1, to account for 1 digit numbers
int yes = 0;
//Determines the size of numArray
while(n/10 != 0)
{
// n%10; <-- What was that all about!
count++;
n = n/10;
}
int numArray[count];
//Fill numArray with each digit of num
for(i = 0; i < count; i++) // This will crash if you use index=count; Array indices go from 0 to Size-1
{
numArray[i] = num%10;
//printf("%d\n", numArray[i]);
num = num/10;
}
//Determines if num is a Palindrome
/*
while(numArray[k] == numArray[count-1]) // Again count-1 not count; This is really bad though what if you have 111111 or some number longer than 6. It might also go out of bounds
{
k = k + 1;
count = count - 1;
yes++;
}
*/
for(k = 1; k <= count; k++)
{
if(numArray[k-1] != numArray[count-k])
return false;
}
return true;
}//End of Function
That's all I could find.
You also need to change this
if(isPalindrome(result) == 0)
To
if(isPalindrome(result))
The Code's Output after making the modifications: Link
The correct printf is the one after the for,after you iterate through all possible values
You are using int to store the value of the palidrome but your result is bigger then 65536,
you should use unsigned
result = i * j;
this pice of code is wrong :
while(n/10 != 0) {
n%10;
count++;
n = n/10;
}
it should be:
while(n != 0) {
count++;
n = n/10;
}
As well as the changes that P.R sugested.
You could do something like this to find out if the number is palindrome:
int isPalindrom(unsigned nr) {
int i, len;
char str[10];
//convert number to string
sprintf(str, "%d", nr);
len = strlen(str);
//compare first half of the digits with the second half
// stop if you find two digits which are not equal
for(i = 0; i < len / 2 && str[i] == str[len - i - 1]; i++);
return i == len / 2;
}
I converted the number to String so I'd be able to go over the number as a char array:
private static boolean isPalindrom(long num) {
String numAsStr = String.valueOf(num);
char[] charArray = numAsStr.toCharArray();
int length = charArray.length;
for (int i = 0 ; i < length/2 ; ++i) {
if (charArray[i] != charArray[length - 1 - i]) return false;
}
return true;
}
I got correct answer for the problem, but I want to know is my coding style is good or bad from my solution. I need to know how can I improve my coding,if it is bad and more generic way.
#include<stdio.h>
#define MAX 999
#define START 100
int main()
{
int i,j,current,n,prev = 0;
for(i = START;i<=MAX;i++)
{
for(j=START;j<=MAX;j++)
{
current = j * i;
if(current > prev) /*check the current value so that if it is less need not go further*/
{
n = palindrome(current);
if (n == 1)
{
printf("The palindrome number is : %d\n",current);
prev = current; // previous value is updated if this the best possible value.
}
}
}
}
}
int palindrome(int num)
{
int a[6],temp;
temp = num;
/*We need a array to store each element*/
a[5] = temp % 10;
a[4] = (temp/10) %10;
a[3] = (temp/100) %10;
a[2] = (temp/1000) %10;
a[1] = (temp/10000) %10;
if(temp/100000 == 0)
{
a[0] = 0;
if(a[1] == a[5] && a[2] == a[4])
return 1;
}
else
{
a[0] = (temp/100000) %10;
if(a[0] == a[5] && a[1] == a[4] && a[2] == a[3])
return 1;
else
return 0;
}
}
No, the largest should be: 906609.
Here is my program:
def reverse_function(x):
return x[::-1]
def palindrome():
largest_num = max(i * x
for i in range(100, 1000)
for x in range(100, 1000)
if str(i * x) == reverse_function(str(i * x)))
return str(largest_num)
print(palindrome())
Related
I'm trying to create a function that compares two four digit numbers and
returns the number of similar digits between the two. For example, with a generated number of 4311 and the user entered 1488,
the score should return 2 (4 and 1).
If it was 4311 and the other is 1147,
the score should return three (1, 1 and 4). I don't know why it isn't giving me the right outputs, hope you can help.
int getSameDigitScore(int playerGuess, int generatedNum) {
int score = 0;
int i;
int j;
int k;
int generatedNumArray[4];
int playerGuessArray[4];
// turns playerGuess into an array
while (playerGuess > 0 ) {
i = 0;
playerGuessArray[i] = playerGuess % 10;
i++;
playerGuess /= 10;
}
// turns generatedNum into an array
while (generatedNum > 0) {
i = 0;
generatedNumArray[i] = generatedNum % 10;
i++;
generatedNum /= 10;
}
// compares the two arrays
for (k = 3; k >= 0; k--) {
for (j = 3; j >= 0; j--) {
if (generatedNumArray[k] == playerGuessArray[j]) {
score++;
playerGuessArray[j] = 0;
j = -5;
}
}
}
return score;
}
You are assigning i = 0 inside the while loop while generating the playerGuessArray and generatedNumArray. Due to which the playerGuess and generatedNumArray array will have elements as first digit of your number 0 0 0 .
Move the initialization out of the loop.
int getSameDigitScore(int playerGuess, int generatedNum) {
int score = 0;
int i, j, k, n;
int generatedNumArray[4];
int playerGuessArray[4];
// turns playerGuess into an array
i = 0; // This has been out of while loop
while (playerGuess > 0 ) {
playerGuessArray[i] = playerGuess % 10;
i++;
playerGuess /= 10;
}
// turns generatedNum into an array
int n = 0; // This has been out of the while loop
while (generatedNum > 0) {
generatedNumArray[n] = generatedNum % 10;
n++;
generatedNum /= 10;
}
// compares the two arrays
for (k = 3; k >= 0; k--) {
for (j = 3; j >= 0; j--) {
if (generatedNumArray[k] == playerGuessArray[j]) {
score++;
playerGuessArray[j] = 0;
j = -5;
}
}
}
return score;
}
int main() {
int m;
n = getSameDigitScore(1231, 2342);
printf("Score is: %d\n", m);
}
You're re-initializing increment variable i on every iteration which should be moved out of the while loop. With that moved out the above code works fine.
There are the following issues with the code.
You are initializing the integer i inside the while loop. This needs to be done before the loop for each loop.
You need a separate array to get the output of equal digits. See AnswerArray in code below. Also it is a good design practice to pass this array to the function and clear this array inside the function.
In the last for loop, you should break from the inner loop after getting a match. This is to take care of cases where playerGuess == 1222 and generatedNum = 1111 In the code shown this will result in a score of 1.
See the final code below with some test cases.
int getSameDigitScore(int playerGuess, int generatedNum, int *AnswerArray) {
int score = 0;
int i;
int j;
int k;
int generatedNumArray[4] = {0};
int playerGuessArray[4] = {0};
memset(AnswerArray,0,4*sizeof(int));
// turns playerGuess into an array
i = 0;
while (playerGuess > 0 ) {
playerGuessArray[i] = playerGuess % 10;
i++;
playerGuess /= 10;
}
// turns generatedNum into an array
i = 0;
while (generatedNum > 0) {
generatedNumArray[i] = generatedNum % 10;
i++;
generatedNum /= 10;
}
// compares the two arrays
score=0;
for (k = 3; k >= 0; k--) {
for (j = 3; j >= 0; j--) {
if (generatedNumArray[k] == playerGuessArray[j]) {
AnswerArray[score++] = generatedNumArray[k];
playerGuessArray[j] = -1;
break;
}
}
}
return score;
}
int main(void)
{
int AnswerArray[4],score;
score = getSameDigitScore(4311,1488,AnswerArray);
printf ("\nScore = %d \n Answer Array = ",score);
for (int i=0; i<score; i++)
{
printf ("%d ",AnswerArray[i]);
}
score = getSameDigitScore(4311,1147,AnswerArray);
printf ("\nScore = %d \n Answer Array = ",score);
for (int i=0; i<score; i++)
{
printf ("%d ",AnswerArray[i]);
}
score = getSameDigitScore(1222,1111,AnswerArray);
printf ("\nScore = %d \n Answer Array = ",score);
for (int i=0; i<score; i++)
{
printf ("%d ",AnswerArray[i]);
}
score = getSameDigitScore(1111,1222,AnswerArray);
printf ("\nScore = %d \n Answer Array = ",score);
for (int i=0; i<score; i++)
{
printf ("%d ",AnswerArray[i]);
}
}
The initializing i=0 which you made inside the loop should be outside the loop.
while (playerGuess > 0 ) {
i = 0;
playerGuessArray[i] = playerGuess % 10;
i++;
playerGuess /= 10;
}
If the initialization is inside the looop then,
Everytime playerGuessArray[0] value will be updated.
FYI:
If playerGuess can contain 0 aat the begin of four digit like 0123
For example, playerGuessValue is 0123, Then by using
while (playerGuess > 0 ) {
i = 0;
playerGuessArray[i] = playerGuess % 10;
i++;
playerGuess /= 10;
}
playerGuessArray will contain only [1,2,3] instead of [0,1,2,3].
So, the better solution would be taking two temporary variables and checking last digit one by one.
Like this:
int temp1=playerGuess, temp2=GeneratedNum;
int i=0;
bool flag = true;
while(flag && i < 4){
if(temp1%10 != temp2%10){
flag = false;
}
temp1 /= 10;
temp2 /= 10;
i++;
}
if(flag){
score++;
}
FYI:
Debugging will help you in finding out these little mistakes.So, try to debug your code with multiple inputs and verify your answer.
Here are few reference on how to debug:
https://blog.hartleybrody.com/debugging-code-beginner/
https://www.codementor.io/mattgoldspink/how-to-debug-code-efficiently-and-effectively-du107u9jh%60
Thanks.
This code is supposed to find the smallest odd number in given array and store it in min but when I try to print min it always prints 0.
int smallestodd(int x[5]){
int j;
int k[5];
int p = 0;
int r = 0;
for(int h =0; h<5;h++){
j = x[h] % 2;
if(j == 1){
int temp =x[h];
k[p] =temp;
p++;
}
}
int min = k[0];
while(k[r] !=0){
if(k[r] < min ){
min = k[r];
r++;
}
}
return min;
}
Assuming there is an odd number in the array -- let's say trying to find the minimum odd number in an array with just even numbers (or no numbers) is UB :)
index = 0;
while (arr[index] % 2 == 0) index++; // skip even numbers
min = arr[index++]; // first odd number
while (index < length) {
if (arr[index] % 2) {
if (arr[index] < min) min = arr[index];
}
index++;
}
this code avoid overflow in search and return 1 when found or 0 if array has only even numbers.
int getMinOdd(int arr[], int length, int *value) {
int found = 0;
for(int idx=0; idx < length; idx++) {
if (arr[idx] % 2) {
if (!found || *value > arr[idx]) {
*value = arr[idx];
}
found = 1;
}
}
return found;
}
It's quite simple actually. You need to just check 2 conditions on your array.
int smallestOdd(int arr[]){
int min = 99999; //Some very large number
for(int i = 0; i < (length of your array); i++) {
if(arr[i]%2 != 0 && arr[i] < min) { //Check if number is odd and less than the current minimum value
min = arr[i];
}
}
return min;
}
Use this using statement as first :
Using System Linq;
Console.WriteLine(myArray.Where(i => i%2 == 1).Min());
I posted earlier about a runtime error in my C program but now I'm having another issue with the code. My program runs fine without any errors but it always prints 0 no matter what the input. I've spent the last 4 hours trying to figure out why my code is doing that but I've had no luck. I would really appreciate if someone could give me a hand.
My program takes in an input of a sequence of integers, using another program that works like scanf. Getint() reads an input of a sequence of integers and stops reading the input when it reaches EOF (-1). The number of integers in the sequence is 1000.
// ar_max(a[]) returns the max entry of a
int ar_max(int a[]) {
int max_so_far = a[0];
for (int i = 1; i < 1000; i++) {
if (a[i] > max_so_far) {
max_so_far = a[i];
}
}
return max_so_far;
}
int main() {
int inputnum = getint();
// array containing the distinct numbers seen
int a_num[1000] = {};
// array containing the frequencies of the distinct numbers seen
int a_freq[1000] = {};
int len_n = 0;
while (inputnum != EOF) {
int i = 0;
len_n = i + 1;
int len_f = len_n;
// update the frequency of inputnum if it's already been seen
for (i = 0; i < len_f; i++, len_n += 1) {
if (a_num[i] == inputnum) {
a_freq[i] = a_freq[i] + 1;
}
}
// add inputnum into the array if it hasn't already been seen
if (i == len_n) {
a_num[i+1] = inputnum;
a_freq[i+1] = 1;
}
inputnum = getint();
}
// print the first number with the highest frequency
for (int j = 0; j < len_n; j++) {
if (a_freq[j] == ar_max(a_freq)) {
printf("%d\n", a_num[j]);
break;
}
}
}
For example, an input of
10 20 30 20
should result in 20
The code has the right idea, but it still has some problems.
First, you have two arrays with related information. You keep two lengths for each of these arrays and try to keep the lengths the same. That's complicated. Instead, consider having just one length for both arrays.
When you loop thorugh the array in oder to check whether the given number is already contained in the array:
for (i = 0; i < len_f; i++, len_n += 1) {
if (a_num[i] == inputnum) {
a_freq[i] = a_freq[i] + 1;
}
}
you update then length of the number array with each iteration. You shouldn't, because you are not changing the length of anything while iterating though the loop.
Next, you check whether the number hasn't been found with:
if (i == len_n) ...
That doesn't work, because you update the frequency in the first loop, but don't terminate the loop; your condition will always be true. You can fix this by breaking out of the loop explicitly with break when you have found the number. (Better yet, make the look-up and frequency update a function whose return value indicates whether the element was found and add the elemen if it wasn't.)
When you append the new element:
if (i == len_n) {
a_num[i+1] = inputnum;
a_freq[i+1] = 1;
}
you should, of course, increment the length of the array you append to. In general, appending an item to an array looks like this:
array[len++] = item;
Recall that the first index is zero and that the actual length is one beyond the valid indices.
There are other points:
When you add an element, you should make sure that you don't overflow the array. The dimension of 1000 is generous, but not infinite.
In your function ar_max, you iterate over all 1000 elements of the array. You have initialised the arrays to zero, so the maximum will still be right, but it is better to pass the array size to the function, too, so that you can iterate over only the actual items.
If your function ar_max returned an index instead of a value, you could access both array with that index and you wouldn't need the last loop over j, because j is just that index.
If you have two parallel arrays, it is a good idea to create a struct that hold the item and frequency and always keeps them together. That simplifies the code, because you don't have to keep anything in sync. Such a data layout also lends itself to sorting and filtering other common operations.
Here's a version that implements the fixes described above. It also uses a function to look up an element by value and makes ar_max return an array index instead of a value.
#include <stdlib.h>
#include <stdio.h>
#define MAX 1000
int getint()
{
int x;
if (scanf("%d", &x) < 1) return EOF;
return x;
}
int ar_max(const int a[], int len)
{
if (len == 0) return -1;
int max_so_far = a[0];
int max_index = 0;
for (int i = 1; i < len; i++) {
if (a[i] > max_so_far) {
max_so_far = a[i];
max_index = i;
}
}
return max_index;
}
int ar_find(const int a[], int len, int which)
{
for (int i = 0; i < len; i++) {
if (a[i] == which) return i;
}
return -1;
}
int main()
{
int inputnum = getint();
int a_num[MAX] = { }; // distinct numbers seen
int a_freq[MAX] = { }; // frequencies of the distinct numbers
int len = 0; // actual length of both arrays
while (inputnum != EOF) {
int i = ar_find(a_num, len, inputnum);
if (i < 0) { // append new item
if (len >= MAX) {
fprintf(stderr, "Array size %d exceeded\n", MAX);
exit(1);
}
a_num[len] = inputnum;
a_freq[len] = 1;
len++;
} else { // increment existing item
a_freq[i]++;
}
inputnum = getint();
}
int imax = ar_max(a_freq, len);
if (imax >= 0) {
printf("%d (%d times)\n", a_num[imax], a_freq[imax]);
} else {
puts("Empty input.");
}
return 0;
}
int main()
{
// array containing the distinct numbers seen
int a_num[1000] = {0};
// array containing the frequencies of the distinct numbers seen
int a_freq[1000] = {0};
int len_n = 0;
int i = 0, j = 0;
// Read first entry
int inputnum = getint();
if (EOF == inputnum)
{
printf("No valid number input given\n");
return -1;
}
// Make first entry in a_num and a_freq
a_num[0] = inputnum;
a_freq[0] = 1;
// Update len_n to 1
len_n = 1;
// Get new entry from user and update number array and frequency array
inputnum = getint();
while (inputnum != EOF)
{
// update the frequency of inputnum if it's already been seen
for (i = 0; i < len_n; i++)
{
if (a_num[i] == inputnum)
{
a_freq[i] = a_freq[i] + 1;
break;
}
}
// add inputnum into the array if it hasn't already been seen
if (i == len_n)
{
a_num[i] = inputnum;
a_freq[i] = 1;
// Update len_n
len_n++;
}
// Check if we have already reached 1000
if (1000 == len_n)
{
printf("Reached 1000 entry read\n");
break;
}
// Next entry
inputnum = getint();
}
// print the number with the highest frequency
int max_freq = ar_max(a_freq);
for (j = 0; j < len_n; j++)
{
if (a_freq[j] == max_freq)
{
printf("%d\n", a_num[j]);
break;
}
}
}
NOTE:
1. #define MAX_SIZE 1000 can be used since hardcoded value 1000 is used repeatedly in code. This will help when you have use new size (say 1500) and you need to change only the #define MAX_SIZE 1500, other code remains unchanged.
2. ar_max() function can be modified to give array index where max occurs to avoid the for loop in main to find the array index. e.g.
int ar_max(int a[])
{
int index = 0;
int max_so_far = a[0];
for (int i = 1; i < 1000; i++) {
if (a[i] > max_so_far) {
max_so_far = a[i];
index = i;
}
}
return index;
}
In main():
printf("%d\n", a_num[ar_max(a_freq)]);
I have to return which digit in a number occurs the most frequently ( though not how many times it occurs )
So far I can only get this, I don't know how to isolate the digit, only to show how many times each digit occurs.
#include <stdio.h>
int frequentDigit(int);
int main()
{
frequentDigit(123032333);
return 0;
}
int frequentDigit(int arg)
{
int tmp; int i; int myArr[9] = { 0 };
tmp = (arg < 0) ? -arg : arg;
do
{
myArr[tmp % 10]++;
tmp /= 10;
} while (tmp != 0);
for (i = 0; i < 9; i++) { printf("\nThere are %d occurances of digit %d", myArr[i], i); }
}
The array where you are storing the frequency of the digits, i.e myArr[]. Its suppose to hold the frequency of all the number from 0...9. And since there are 10 numbers, you would need an array of lenght 10.
int myArr[10];
Later, you need to traverse through the array once, checking for the max element, and saving the index accordingly, to find which number has occured most number of times.
To traverse, the for loop should go till 9
for (i = 0; i <= 9; i++)
Edited
As someone commented, you can find the max value while you are computing the frequencies itself.
int max = -1, max_num = -1;
do
{
myArr[tmp % 10]++;
if( myArr[tmp % 10] > max)
{
max = myArr[tmp % 10];
max_num = tmp % 10;
}
tmp /= 10;
} while (tmp != 0);
printf("%d", max_num);
Its simple. At the end of your code you have an array of frequencies, if you find the max of that you get the most common element
Just use a loop to find the max and print that:
int max = myArr[0]; // start with max = first element
int max_position=0;
for(int i = 1; i<9; i++)
{
if(myArr[i] > max){
max = myArr[i];
max_position=i;
}
}
printf("\The max is %d occuring %d times ", max_position, max_position)
I have this assignment for my intro to C programming class and part of my code has to find the sequence of the sum of square digits of a number in order to determine after if the given number is a happy number (sum of square digits = 1)
Here's part of my code:
#include <stdio.h>
#include <math.h>
// The sum of square digits function
int sqd (int x) {
int sum = 0;
while (x > 0) {
sum = sum + pow(x%10, 2);
x = x/10;
}
return sum;
}
// The search function
int search (int a[], int val, int size) {
int i;
for (i = 0; i < size; i++) {
if (a[i] == val) {
return 1;
}
}
return 0;
}
// The main program
void main () {
int a [1000] = {0};
int N;
int count = 1;
int j;
printf("Please enter the potential happy number:\n", N);
scanf ("%d", &N);
a[0] = N;
a[count] = sqd (N);
do {
a[count] = sqd (a[count-1]);
count++;
} while (search (a, a[count], count));
for ( j = 0; j <= count; j++) {
printf("%d\n", a[j]);
}
}
It only prints the first three sums in the sequence. I really don't know how to make it work.
Thank you in advance
This line
while (search (a, a[count], count));
makes sure that you break out of the loop after one round since a[1] is not equal toa[0]. You can change that line to be:
while (a[count-1] != 1);
You also need to add a clause to make sure that you stop when the limit of the array is reached. Update that line to be:
while (a[count-1] != 1 && count < 1000 );
And then, change the printing loop to use i < count, not i <= count. Using <= will result in accessing the array out of bounds when the user enters a sad number.
for ( j = 0; j < count; j++){
printf("%d\n", a[j]);
}
Update
After a bit of reading on happy numbers at Wikipedia, I understand why you had call to search in the conditional of the while. The following also works.
} while ( ! (a[count-1] == 1 || search(a, a[count-1], count-1)) );
That will search for the last number in the array but only up to the previous index.