I have an assignment to write a program for a natural number where its inverse is divisible by its number of digits. A natural number n ( n > 9) is entered from the keyboard. To find and print the largest natural number less than n that its inverse is divisible by its number of digits. If the entered number is not valid, print a corresponding message (Brojot ne e validen).
I have tried :
#include <stdio.h>
int main() {
int n,r,s=0,a=0;
int m;
scanf("%d",&n);
int t=n;
if(t<10)
{ printf("Brojot ne e validen");}
else {
for (int i = n - 1; i > 0; i--) {
while (n != 0) {
r = n % 10;
s = (s * 10) + r;
n = n / 10;
a++;
if (s % a == 0) {
m = i;
break;
}
}
}
printf("%d\n", m);
}
return 0;
}
And when my inputs is 50, it gives the correct answer which is 49, but when I try numbers like 100 or 17 it prints 98 instead of 89 and 16 instead of 7 respectively. I have been baffled by this for more than an hour now
check your logic.
you can check each value by
#include <stdio.h>
int main() {
int t,r,s=0,a=0;
int m;
scanf("%d",&t);
if(t<10)
{ printf("Brojot ne e validen");}
else {
for (int i = t - 1; i > 0; i--) {
while (t != 0) {
r = t % 10;
printf("%d \n", r);
s = (s * 10) + r;
printf("%d \n", s);
t = t / 10;
printf("%d \n", t);
a++;
if (s % a == 0) {
m = i;
break;
}
}
}
printf("%d\n", m);
}
return 0;
}
Related
A number and a reversed number form a pair. If both numbers are prime numbers, we call it a reversed prime number pair. For instance, 13 and 31 is a 2 digit reversed prime number pair, 107 and 701 is a 3 digit reversed prime number pairs.
Write a program to output all n (2<=n<=5) digit reversed prime number pairs. If the input is less than 2 or greater than 5, output "Wrong input." and terminate the program. While ouputting , every 5 pairs form a new line, and only output the pair in which the first number is smaller than the second number.
Input: 1
Output: Wrong input.
Input: 3
Output:
(107,701)(113,311)(149,941)(157,751)(167,761)
(179,971)(199,991)(337,733)(347,743)(359,953)
(389,983)(709,907)(739,937)(769,967)
There are 14 results.
Can anyone give me hints how to do this?
I know how to determine if a number is a reversed prime number, but i couldn't understand how to complete this challenge from my friend
#include <stdio.h>
int checkPrime(int n) {
int i, isPrime = 1;
if (n == 0 || n == 1) {
isPrime = 0;
}
else {
for(i = 2; i <= n/2; ++i) {
if(n % i == 0) {
isPrime = 0;
break;
}
}
}
return isPrime;
}
int main (void)
{
int a, reverse = 0, remainder, flag=0;
scanf("%d",&a);
int temp = a;
while (temp!=0) {
remainder = temp%10;
reverse = reverse*10 + remainder;
temp/=10;
}
if (checkPrime(a)==1) {
if (checkPrime(reverse)==1){
printf("YES\n");
flag=1;
}
}
if (flag==0)
printf("NO\n");
}
This will be the correct solution:
#include <stdio.h>
#include <stdbool.h>
#include <math.h>
#include <stdlib.h>
#define MAX_N 100000
int *primes;
int num_primes;
void init_primes() {
int sqrt_max_n = sqrt(MAX_N);
primes = (int *) malloc(sqrt_max_n / 2 * sizeof(int));
num_primes = 0;
primes[num_primes] = 2;
num_primes++;
for (int i = 3; i <= sqrt_max_n; i += 2) {
bool is_prime = true;
for (int j = 0; j < num_primes; j++) {
if (i % primes[j] == 0) {
is_prime = false;
break;
}
}
if (is_prime) {
primes[num_primes] = i;
num_primes++;
}
}
}
int is_prime(int n) {
for (int i = 0; i < num_primes; i++) {
if (primes[i] == n) {
return 1;
}
if (n % primes[i] == 0) {
return 0;
}
}
return 1;
}
int reverse(int n) {
int reversed_n = 0;
while (n > 0) {
reversed_n = reversed_n * 10 + n % 10;
n /= 10;
}
return reversed_n;
}
int main() {
init_primes();
int n;
printf("Enter n (2 <= n <= 5): ");
scanf("%d", &n);
if (n < 2 || n > 5) {
printf("Wrong input.\n");
return 0;
}
int min = (int) pow(10, n - 1);
int max = (int) pow(10, n) - 1;
int count = 0;
for (int i = min; i <= max; i++) {
if (is_prime(i)) {
int reversed_i = reverse(i);
if (i < reversed_i && is_prime(reversed_i)) {
printf("(%d %d)", i, reversed_i);
count++;
if (count % 5 == 0) {
printf("\n");
} else {
printf(" ");
}
}
}
}
return 0;
}
After testing this code I get the same result what you need:
Enter n (2 <= n <= 5): 3
(107 701) (113 311) (149 941) (157 751) (167 761)
(179 971) (199 991) (337 733) (347 743) (359 953)
(389 983) (709 907) (739 937) (769 967)
The init_primes method caches all the required prime numbers until the sqrt of your limit to a dynamic array.
The is_prime method uses that cache for detecting whether a number is prime or not.
I am trying to print the series but whenever I set the range (input given by me) above 407. I only get the output till 407. However, when I set the range below 407 it gives me the result according to the input I have given. Can anybody tell me what I'm doing wrong?
I used an online compiler (www.onlinegdb.com) to write my code.
Here is the code.
#include<stdio.h>
#include<stdlib.h>
int
main ()
{
int m, n;
printf
("Enter two numbers to find the Armstrong numbers that lie between them.\n");
scanf ("%d%d", &m, &n);
system("clear");
if(m>n)
{
m = m + n;
n = m - n;
m = m - n;
}
for (; m < n; m++)
{
int i = m + 1, r, s = 0, t;
t = i;
while (i > 0)
{
r = i % 10;
s = s + (r * r * r);
i = i / 10;
}
if (t == s)
printf ("%d ", t);
}
return 0;
}
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Try this code!!!
#include <math.h>
#include <stdio.h>
int main() {
int low, high, number, originalNumber, rem, count = 0;
double result = 0.0;
printf("Enter two numbers(intervals): ");
scanf("%d %d", &low, &high);
printf("Armstrong numbers between %d and %d are: ", low, high);
// swap numbers if high < low
if (high < low) {
high += low;
low = high - low;
high -= low;
}
// iterate number from (low + 1) to (high - 1)
// In each iteration, check if number is Armstrong
for (number = low + 1; number < high; ++number) {
originalNumber = number;
// number of digits calculation
while (originalNumber != 0) {
originalNumber /= 10;
++count;
}
originalNumber = number;
// result contains sum of nth power of individual digits
while (originalNumber != 0) {
rem = originalNumber % 10;
result += pow(rem, count);
originalNumber /= 10;
}
// check if number is equal to the sum of nth power of individual digits
if ((int)result == number) {
printf("%d ", number);
}
// resetting the values
count = 0;
result = 0;
}
return 0;
}
Try this code :
#include <stdio.h>
#include <math.h>
int main()
{
int start, end, i, temp1, temp2, remainder, n = 0, result = 0;
printf(“Enter start value and end value : “);
scanf(“%d %d”, &start, &end);
printf(“\nArmstrong numbers between %d an %d are: “, start, end);
for(i = start + 1; i < end; ++i)
{
temp2 = i;
temp1 = i;
while (temp1 != 0)
{
temp1 /= 10;
++n;
}
while (temp2 != 0)
{
remainder = temp2 % 10;
result += pow(remainder, n);
temp2 /= 10;
}
if (result == i) {
printf(“%d “, i);
}
n = 0;
result = 0;
}
printf(“\n”);
return 0;
}
1
2 4
3 5 7
6 8 10 12
9 11 13 15 17
Following is the code in which I am not able to print the pyramid:-
int main()
{
int i,j;
for(i=1;i<=5;i++){
for(j=1;j<=i;j++){
printf("%d ",i*j);
}
printf("\n");
}
return 0;
}
You need to track both even and odd numbers .
#include <stdio.h>
int main()
{
int even=1,odd=2;
int n=10;
for (int i = 1; i <= n; i++)
{
int a= (i % 2 == 0);
for (int j = 1; j < i; j++)
{
if(a)
{
printf("%d ",even);
}
else
{
printf("%d ",odd);
}
even += a ? 2 : 0;
odd += a ? 0 : 2;
}
printf("\n");
}
return 0;
}
Not very clean and compact algorithm but sth like this would work:
#include <stdio.h>
#include <stdlib.h>
int main() {
char tmp[10];
int n = 0, row = 1, odd = 1, even = 2, c = 0, selectOdd, fin = 0;
printf("maximum number: ");
scanf("%s", tmp);
n = atoi(tmp);
if (n != 0) {
while (fin < 2) {
selectOdd = row % 2;
c = row;
if (selectOdd) {
while (c != 0) {
printf("%3d", odd);
odd += 2;
if (odd > n) {
fin++;
break;
}
c--;
}
}
else {
while (c != 0) {
printf("%3d", even);
even += 2;
if (even > n) {
fin++;
break;
}
c--;
}
}
printf("\n");
row++;
}
}
return 0;
}
it's simple
your algorithm is odd, even, odd,... and so on
so you start with odd number until reach line number
for next line is even and you can find start number with this
you just need find number at start of line and continue print number number
in each step you just need
num += 2;
remember 'lineIndex' start from 1
num = (lineIndex - 1) * 2 + lineIndex % 2;
this is a full code
#include <stdio.h>
int main(){
int numIndex;
int lineIndex;
int num;
for (lineIndex = 1; lineIndex <= 5; lineIndex++) {
num = (lineIndex - 1) * 2 + lineIndex % 2;
for (numIndex = 0; numIndex < lineIndex; numIndex++) {
printf("%2d ", num);
num += 2;
}
printf("\n");
}
}
I have a code that finds the sum of the divisors of a number, but I can't get it to apply on my increasing n and print all the numbers respectively.
The code is
long div(int n) {
long sum = 0;
int square_root = sqrt(n);
for (int i = 1; i <= square_root; i++) {
if (n % i == 0) {
sum += i;
if (i * i != n) {
sum += n / i;
}
}
}
return sum - n;
}
On my main() I need to have a c number that starts from 1 and goes to my MAXCYC which is 28. The n goes from 2 to MAXNUM which is 10000000. The program needs to find all perfect, amicable and sociable numbers and print them with their respective pairs.
Sample output:
Cycle of length 2: 12285 14595 12285
Cycle of length 5: 12496 14288 15472 14536 14264 12496
for (int n = 2; n <= MAXNUM; n++) {
long sum = div(n);
long res = div(sum);
if (res <= MAXNUM) { // Checking if the number is just sociable
int c = 0;
while (c <= MAXCYC && n != res) {
res = div(sum);
c++;
}
if (c <= MAXCYC) {
printf("Cycle of length %d: ", c);
printf("%ld ", sum);
do {
printf("%ld ", res);
res = div(res);
}
while (sum < res);
printf("%ld ", sum);
c += c - 2;
printf("\n");
}
}
}
I only get pairs of cycle length of 1, 2 and nothing above that. Also it doesn't even print it correctly since it says Cycle of length 0: in all of the results without increasing. I think the problem is in the f before the first print but I can't get it to work in a way that as long as my
(n == sum) it prints Cycle of length 1: x x pairs
(n == res && sum < res) it prints Cycle of length 2: x y x pairs
(res <= MAXNUM) it prints Cycle of length c: x y z ... x (c amount of pairs including first x)
What do you guys think I should change?
Ok, this code should work if I understood well your requirement.
#include <stdio.h>
#include <stdlib.h>
int div_sum(int n)
{
long sum = 0;
int square_root = sqrt(n);
for (int i = 1; i <= square_root; i++)
{
if (n % i == 0)
{
sum += i;
if (i * i != n)
{
sum += n / i;
}
}
}
return sum - n;
}
int MAX_N = 10000000;
int MAX_CYCLES = 28;
int main()
{
int cycles;
for(int n = 2; n < MAX_N; n++){
int found = 0;
for(int c = 1; !found && c <= MAX_CYCLES; c++){
cycles = c;
int aliquote = n;
while(cycles--) aliquote = div_sum(aliquote);
//it is a cycle of length c
cycles = c;
if(n == aliquote){
printf("Cycle of length %d: %d", c, n);
while(cycles--){
aliquote = div_sum(aliquote);
printf(" %d", aliquote);
}
printf("\n");
found = 1;
}
}
}
return 0;
}
I need to find the sum of all numbers that are less or equal with my input number (it requires them to be palindromic in both radix 10 and 2). Here is my code:
#include <stdio.h>
#include <stdlib.h>
int pal10(int n) {
int reverse, x;
x = n;
while (n != 0) {
reverse = reverse * 10 + n % 10;
n = n / 10;
}
if (reverse == x)
return 1;
else
return 0;
}
int length(int n) {
int l = 0;
while (n != 0) {
n = n / 2;
l++;
}
return l;
}
int binarypal(int n) {
int v[length(n)], i = 0, j = length(n);
while (n != 0) {
v[i] = n % 2;
n = n / 2;
i++;
}
for (i = 0; i <= length(n); i++) {
if (v[i] == v[j]) {
j--;
} else {
break;
return 0;
}
}
return 1;
}
int main() {
long s = 0;
int n;
printf("Input your number \n");
scanf("%d", &n);
while (n != 0) {
if (binarypal(n) == 1 && pal10(n) == 1)
s = s + n;
n--;
}
printf("Your sum is %ld", s);
return 0;
}
It always returns 0. My guess is I've done something wrong in the binarypal function. What should I do?
You have multiple problems:
function pal10() fails because reverse is not initialized.
function binarypal() is too complicated, you should use the same method as pal10().
you should avoid comparing boolean function return values with 1, the convention in C is to return 0 for false and non zero for true.
you should avoid using l for a variable name as it looks very similar to 1 on most constant width fonts. As a matter of fact, it is the same glyph for the original Courier typewriter font.
Here is a simplified and corrected version with a multi-base function:
#include <stdio.h>
#include <stdlib.h>
int ispal(int n, int base) {
int reverse = 0, x = n;
while (n > 0) {
reverse = reverse * base + n % base;
n = n / base;
}
return reverse == x;
}
int main(void) {
long s = 0;
int n = 0;
printf("Input your number:\n");
scanf("%d", &n);
while (n > 0) {
if (ispal(n, 10) && ispal(n, 2))
s += n;
n--;
}
printf("Your sum is %ld\n", s);
return 0;
}
in the function pal10 the variable reverse is not initialized.
int pal10(int n)
{
int reverse,x;
^^^^^^^
x=n;
while(n!=0)
{
reverse=reverse*10+n%10;
n=n/10;
}
if(reverse==x)
return 1;
else
return 0;
}
In the function binarypal this loop is incorrect because the valid range of indices of an array with length( n ) elements is [0, length( n ) - 1 ]
for(i=0;i<=length(n);i++)
{
if(v[i]==v[j])
{
j--;
}
else
{
break;
return 0;
}
}
And as #BLUEPIXY pointed out you shall remove the break statement from this else
else
{
break;
return 0;
}