I want do a program of factorization like this: 72 = 3 * 2 ^ 3 ^ 2
in C language, how can i do the program ?
I tried to do this but I can not do it :
#include <stdio.h>
int main(){
int n;
int j;
printf("Insert a positive integer number greater than 1\n")
scanf("%d", &n);
j = 2;
do
{
if( n % j == 0)
{
printf("%d\n", j);
n = n / j;
}
else{
j++;
}
}
while ( n > 1);
}
Assuming that given 72, you want to output 2^3 X 3^2 this code should do it:
/* Decides when to print multiplication sign */
const char *get_mult_string()
{
static int first_divisor=1;
if (first_divisor==1) {
first_divisor=0;
return "";
} else {
return " X ";
}
}
void factorize() {
int n;
int j;
printf("Insert a positive integer number greater than 1: ");
scanf("%d", &n);
j = 2;
int power_count=0;
do {
if (n % j == 0) {
power_count++;
n = n / j;
}
else {
if (power_count>0) {
printf("%s(%d^%d)", get_mult_string(), j, power_count);
power_count=0;
}
j++;
}
}
while (n > 1);
if (power_count>0) {
printf("%s(%d^%d)\n", get_mult_string(), j, power_count);
}
}
Try creating an array, where each slot represents a prime number, and initialize them all to 0. For example, array[0] means 2, array[1] 3, array[2] 5, array[3] 7, array[4] 11.
Rather than printing out the prime number when you reach it, as you do in your code, increment that slot in the array. In the end, you can call something like printf("2^%d * 3^%d * 5^%d * 7^%d * 11^%d",array[0],array[1],array[2],array[3],array[4]).
Note: switch statements would probably be your friend here.
Related
I need to write a program that allows user to enter an array of integers, finds the digit that appears most often in all entered numbers, and removes it from the elements of the array. If several digits appear the same number of times, the smallest of them should be deleted. If all digits of the element of the array are deleted, that element should become zero. In the end, such a modified array is printed.
Example of input and output:
Enter number of elements of the array: 5
Enter the array: 3833 8818 23 33 1288
After deleting, the array is: 8 8818 2 0 1288
Explanation: The numbers 3 and 8 appear the same number of times (6 times each), but 3 is less, so it was removed it from all members of the array. Element 33 consists exclusively of the digits 3, so that it becomes 0.
#include <stdio.h>
int main() {
int i,n,arr[100]; n;
printf("Enter number of elements of the array: ");
scanf("%d", &n);
printf("Enter the array: ");
for(i=0;i<n;i++) {
scanf("%d", &arr[i]);
}
return 0;
}
EDIT: I'm beginner to programming, and this task should be done using only knowledge learned so far in my course which is conditionals, loops, and arrays. This shouldn't be done with strings.
Divide the problem into separate tasks.
Write the code
In the code below I do not treat 0 as having digit 0. It is because it is not possible to remove 0 from 0. You can easily change this behaviour by changing while(){} loop to do{}while()
int removeDigit(int val, int digit)
{
int result = 0;
unsigned mul = 1;
int sign = val < 0 ? -1 : 1;
digit %= 10;
while(val)
{
int dg = abs(val % 10);
if(dg != digit)
{
result += dg * mul;
mul *= 10;
}
val /= 10;
}
return sign * result;
}
void countDigits(int val, size_t *freq)
{
while(val)
{
freq[abs(val % 10)]++;
val /= 10;
}
}
int findMostFrequent(const size_t *freq)
{
size_t max = 0;
for(size_t i = 1; i < 10; i++)
{
if(freq[i] > freq[max]) max = i;
}
return (int)max;
}
int main(void)
{
int table[20];
size_t freq[10] = {0,};
int mostfreq = 0;
srand(time(NULL));
for(size_t i = 0; i < 20; i++)
{
table[i] = rand();
printf("Table[%zu] = %d\n", i, table[i]);
countDigits(table[i], freq);
}
mostfreq = findMostFrequent(freq);
printf("Most frequent digit: %d\n", mostfreq);
for(size_t i = 0; i < 20; i++)
{
table[i] = removeDigit(table[i], mostfreq);
printf("Table[%zu] = %d\n", i, table[i]);
}
}
https://godbolt.org/z/PPj9s341b
I wrote this program to find prime numbers between 1 and 50000, and I still need to find how many prime numbers there is (I tried a lot of tricks but I did not succeed)
#include <stdio.h>
//int getValueFromUser();
void PrintListOfPrime(int value);
int main() {
int value = 23;
PrintListOfPrime(value);
return 0;
}
void PrintListOfPrime(int value) {
int ValueIsPrime; //ValueIsPrime is used as flag variable
printf("The list of primes: ");
for (int i = 2; i <= value; i++) {
ValueIsPrime = 1;
/* Check if the current number i is prime or not */
for (int j = 2; j <= i / 2; j++) {
/*
* If the number is divisible by any number
* other than 1 and self then it is not prime
*/
if (i % j == 0) {
ValueIsPrime = 0;
break;
}
}
/* If the number is prime then print */
if (ValueIsPrime == 1)
printf("%d, ", i);
}
printf("\n");
}
I tried a lot of tricks but I did not succeed
If OP's code takes too long to ran, iterate to the square root of i, not up to i/2.
j <= i / 2 is very slow. Use j <= i / j instead.
Form a count and increment with every prime. #gspr
if (ValueIsPrime == 1) {
printf("%d, ", i);
prime_count++;
}
Bigger change yet even faster to "find prime numbers between 1 and 50000", research Sieve of Eratosthenes
Hello fast answer is to create a variable in main, int totaleOfPrimes = 0; for example.
then send it by reference to the fucntion :
Function declaration : void PrintListOfPrime(int value,int* counter);
Function call : void PrintListOfPrime(value,&totaleOfPrimes);
then Increment counter befor printing :
if (ValueIsPrime == 1){
(*counter)++;
printf("%d, ", i);
}
There is no need to iterate the loops for all numbers between 2 and value. You should consider only 2 and odd numbers.
The function can look the following way as it is shown in the demonstrative program below.
#include <stdio.h>
static inline size_t PrintListOfPrime( unsigned int n )
{
size_t count = 0;
printf( "The list of primes:\n" );
for ( unsigned int i = 2; i <= n; i = i != 2 ? i + 2 : i + 1 )
{
int isPrime = 1;
/* Check if the current number i is prime or not */
for ( unsigned int j = 3; isPrime && j <= i / j; j += 2 )
{
/*
* If the number is divisible by any number
* other than 1 and self then it is not prime
*/
isPrime = i % j != 0;
}
/* If the number is prime then print */
if ( isPrime )
{
if ( ++count % 14 == 0 ) putchar( '\n' );
printf( "%u ", i );
}
}
return count;
}
int main(void)
{
unsigned int n = 50000;
size_t count = PrintListOfPrime( n );
printf( "\n\nThere are %zu prime numbers up to %u\n", count, n );
return 0;
}
Run this code in C. It will return the value of a pi(x) function. It is basically the Prime counting function:
#include <stdio.h>
#define LEAST_PRIME 2
#include <math.h>
int main() //works for first 10000 primes.
{
int lower_limit = 2, no_of_sets;
// printf("NUMBER OF SETS: ");
// scanf("%d", &no_of_sets);
int remainder, divisor = 2, remainder_dump, upper_limit; //upper limit to be specified
//by user.
int i = 1;
// printf("SPECIFY LOWER LIMIT: ");
// scanf("%d", &lower_limit);
int number_to_be_checked = lower_limit;
printf("SPECIFY UPPER LIMIT: ");
scanf("%d", &upper_limit);
printf("2\t\t\t\t", number_to_be_checked);
//PRINTS 2.*/
do
{
remainder_dump = 1;
divisor = 2;
do
{
remainder = number_to_be_checked % divisor;
if (remainder == 0)
{
remainder_dump = remainder_dump * remainder; // dumping 0 for rejection.
break;
}
++divisor;
} while (divisor <= number_to_be_checked / divisor); // upto here we know number
is prime or not.
if (remainder_dump != 0)
{
++i;
printf("%d.\t\t\t\t", number_to_be_checked); //print if prime.
};
number_to_be_checked = number_to_be_checked + 1;
} while (number_to_be_checked <= upper_limit);
printf("\n pi(x) = %d \n", i);
//printf("pi function value is %f.", (i - 1) / (log(i - 1)));
float app;
app = upper_limit / (log(upper_limit));
float plot_value;
plot_value = (i) / app;
printf(" BETA FUNCTION VALUE ~ %f", plot_value);
return 0;
}
I am currently making a program for the collatz sequence on C, however the last value, which is 1, is not being printed. For example, when I input 8, the outcome must be 8 4 2 1, but it only prints 8 4 2, or when I input 5, it only prints 5 16 8 4 2. What can I put inside the while ( ) to print the complete answer? Thank you!!
void
CollatzSequence(int n)
{
int x = 1;
do {
x++;
printf("%3d", n);
if (n%2==0)
n /= 2;
else
n = 3 * n + 1;
}
while ( );
printf("\n");
}
int
main()
{
int n;
do {
printf("Input an integer greater than 0: ");
scanf("%d", &n);
if (n <= 0)
printf("Invalid input. Try again.\n");
} while (n <= 0);
CollatzSequence(n);
return 0;
}
Code needs new loop exit condition.
Loop is done once 1 is printed.
Sample below.
void CollatzSequence(int n) {
for (;;) {
printf("%3d", n);
if (n == 1)
break;
if (n % 2 == 0)
n /= 2;
else
n = 3 * n + 1;
}
printf("\n");
}
I have return the code to find a factorial and to display trailing zeros at the end of the factorial, but the output is wrong... could you please help me to find the mistake?
#include <stdio.h>
int main() {
int m = 1, i, N, count = 0;
scanf("%d", &N);
for (i = 1; i <= N; i++) {
m = m * i;
}
printf("%d", m);
while (m > 0) {
if ((m % 10) == 0) {
count = count + 1;
m = m / 10;
}
break;
}
printf("%d", count);
return 0;
}
Your code only works for very small values of N: up to 9. For slightly larger values, you would need to add an else keyword before the break statement and you would get a correct result for a few more cases.
For larger values, you must compute the power of 5 that divides the factorial. You can do this incrementally by summing the power of 5 that divide each individual number up to and including N.
#include <stdio.h>
int main() {
int N, count;
if (scanf("%d", &N) != 1)
return 1;
/* only consider factors that are multiples of 5 */
count = 0;
for (int i = 5; i <= N; i += 5) {
for (int j = i; j % 5 == 0; j /= 5)
count++;
}
printf("%d\n", count);
return 0;
}
An even simpler and faster solution is this: compute the number of multiples of 5 less or equal to N, add the number of multiples of 5*5, etc.
Here is the code:
#include <stdio.h>
int main() {
int N, count;
if (scanf("%d", &N) != 1)
return 1;
count = 0;
for (int i = N; (i /= 5) > 0;) {
count += i;
}
printf("%d\n", count);
return 0;
}
you have two problems
your collapse the two outputs so you see only one of them / you cannot see who is who, just add a separator between them
an else is missing when you count so you count to only up to 1 and the result is wrong from factorial 10
So the minimal changes produce :
int main()
{
int m=1,i,N,count=0;
scanf("%d",&N);
for(i=1;i<=N;i++)
{
m=m*i;
}
printf("%d\n",m); /* <<< added \n */
while(m>0)
{
if((m%10)==0)
{
count=count+1;
m=m/10;
}
else /* <<< added else */
break;
}
printf("%d\n",count); /* <<< added \n */
return 0;
}
after the changes :
pi#raspberrypi:/tmp $ ./a.out
5
120
1
pi#raspberrypi:/tmp $ ./a.out
10
3628800
2
Of course that supposes first you are able to compute the factorial without overflow
I also encourage you to check a value was read by scanf, checking it returns 1
#include <stdio.h>
int main()
{
int n,i,f=1,t,c=0;
printf("Enter number ");
scanf("%d",&n);
t=n;
for(i=1;t>=5;i++)
{
t=n/5;
c=c+t;
n=t;
}
printf("number of zeros are %d",c);
return 0;
}
Please help me to solve this task:
Generate all binary strings of length n with k bits set.(need to write on C)
for example:
n=5
k=3
11100
00111
11010
01011
**01110
11001
10011
**01101
**10110
10101
** can't generate these permutations
Code:
#include <stdio.h>
#define N 10
int main (void)
{
int mas[N]={0},kst,m,n1,z,a,b;
printf("\n\nVvedit` rozmirnist` masyvu: ");
scanf("%d",&kst);
printf("\n\nVvedit` kil`kist` odynyc`: ");
scanf("%d",&n1);
for(m=0;m1;m++)
mas[m]=1;
for(m=0;m<kst;m++)
printf("%d",mas[m]);
printf("\n");
for(m=0;m<n1;m++){
for(z=0;z<(kst-1);z++)
if((mas[z]==1) && (mas[z+1]==0)){
a=mas[z];
mas[z]=mas[z+1];
mas[z+1]=a;
for(b=0;b<kst;b++)
printf("%d",mas[b]);
printf("\n");
}
}
return 0;
}
I have solved this problem earlier! please find my code below! I hope this will help you out.
#include<stdio.h>
int NumberOfBitsSet(int number)
{
int BitsSet = 0;
while(number != 0)
{
if(number & 0x01)
{
BitsSet++;
}
number = number >> 1;
}
return BitsSet;
}
void PrintNumberInBinary(int number, int NumBits)
{
int val;
val = 1 << NumBits; // here val is the maximum possible number of N bits with only MSB set
while(val != 0)
{
if(number & val)
{
printf("1");
}
else
{
printf("0");
}
val = val >> 1;
}
}
int main()
{
int n,k,i;
int max,min;
printf("enter total number of bits and number of bits to be set:\n");
scanf("%d %d", &n, &k);
min = ((1 << k) - 1); //min possible values with k bits set
max = (min << (n-k)); //max possible value with k bits set!
//printf("%d %d", min, max);
for(i=0; i<= max; i++)
{
if(!(i<min))
{
if(NumberOfBitsSet(i) == k)
{
PrintNumberInBinary(i, (n-1));
printf("\n");
}
}
}
return 0;
}
Your code is a mess ;)
Seriously: first rule when solving a task in code is to write clean code, use sensible variable naming etc.
For tasks like this one I would suggest using this.
Now to your sample code: it would not compile and it is hard to read what you are trying to do. Formatted and with some comments:
#include <stdio.h>
#define N 10
int main(void)
{
int mas[N] = {0};
int kst, m, n1, z, a, b;
/* Read width ? */
printf("\n\nVvedit` rozmirnist` masyvu: ");
scanf("%d", &kst);
/* Read number of bit's set? */
printf("\n\nVvedit` kil`kist` odynyc`: ");
scanf("%d", &n1);
/* m1 is not defined, thus the loop give no meaning.
* Guess you are trying to set "bits" integers to 1.
*/
for (m = 0; m1; m++)
mas[m] = 1;
/* This should be in a function as 1. You do it more then once, and
* 2. It makes the code much cleaner and easy to maintain.
*/
for (m = 0; m < kst; m++)
printf("%d", mas[m]);
printf("\n");
for (m = 0; m < n1; m++) {
for (z = 0; z < (kst - 1); z++) {
if ((mas[z] == 1) && (mas[z + 1] == 0)) {
a = mas[z]; /* Same as a = 1; */
mas[z] = mas[z + 1]; /* Same as mas[z] = 0; */
mas[z + 1] = a; /* Same as mas[z + 1] = 1; */
/* Put this into a function. */
for (b = 0; b < kst; b++)
printf("%d", mas[b]);
printf("\n");
}
}
}
return 0;
}
The extensive use of printf when one are not sure of what is going on is a precious tool.
This is not a solution, (it is basically doing the same as your post, but split up), but a sample of something that might be easier to work with. I have also used a char array as C-string instead of integer array. Easier to work with in this situation.
If you want to use integer array I'd suggest you add a print_perm(int *perm, int width) helper function to get it out of the main code.
#include <stdio.h>
#define MAX_WIDTH 10
int get_spec(int *width, int *bits)
{
fprintf(stderr, "Enter width (max %-2d): ", MAX_WIDTH);
scanf("%d", width);
if (*width > MAX_WIDTH) {
fprintf(stderr, "Bad input: %d > %d\n", *width, MAX_WIDTH);
return 1;
}
fprintf(stderr, "Enter set bits (max %-2d): ", *width);
scanf("%d", bits);
if (*bits > MAX_WIDTH) {
fprintf(stderr, "Bad input: %d > %d\n", *bits, MAX_WIDTH);
return 1;
}
return 0;
}
void permutate(int width, int bits)
{
char perm[MAX_WIDTH + 1];
int i, j;
/* Set "bits" */
for (i = 0; i < width; ++i)
perm[i] = i < bits ? '1' : '0';
/* Terminate C string */
perm[i] = '\0';
fprintf(stderr, "\nPermutations:\n");
printf("%s\n", perm);
for (i = 0; i < bits; ++i) {
/* Debug print current perm and outer iteration number */
printf("%*s LOOP(%d) %s\n",
width, "", i, perm
);
for (j = 0; j < (width - 1); ++j) {
if (perm[j] == '1' && perm[j + 1] == '0') {
perm[j] = '0';
perm[j + 1] = '1';
printf("%s j=%d print\n",
perm, j
);
} else {
/* Debug print */
printf("%*s j=%d skip %s\n",
width, "", j, perm
);
}
}
}
}
int main(void)
{
int width, bits;
if (get_spec(&width, &bits))
return 1;
permutate(width, bits);
return 0;
}
If you want to list all of the permutations uniquely without doing "iterate and check", you can do something like this:
# Move peg x up m using s
# x is negative
# m is positive
def move(x, m, s):
for i in range(1, m+1):
s2 = list(s)
s2[x] = 0
s2[x - i] = 1
print(s2)
if x + 1 < 0:
move(x+1, i, s2)
# Print all unique permutations of
# n bits with k ones (and n-k zeros)
def uniqPerms(n, k):
s = [0 for _ in range(n-k)] + [1 for _ in range(k)]
print(s)
move(-k, n-k, s)
if __name__ == '__main__':
from sys import argv
uniqPerms(int(argv[1]), int(argv[2]))
The idea is that you inch the 1's up recursively, so that each movement produces a unique list (since a 1 is now somewhere none was before).
And you said it must be in C:
#include <stdio.h>
#include <stdlib.h>
enum { n = 8 };
struct string
{
char str[n + 1];
};
void move(int x, int m, string s)
{
for (int i = 0; i <= m; ++i)
{
string s2 = s;
s2.str[n + x] = '0';
s2.str[n + x - i] = '1';
printf("%s\n", s2.str);
if (x + 1 < 0)
move(x + 1, i, s2);
}
}
void uniqPerms(int k)
{
string s;
for (int i = 0; i < n - k; ++i)
s.str[i] = '0';
for (int i = n - k; i < n; ++i)
s.str[i] = '1';
s.str[n] = '\0';
printf("%s\n", s.str);
move(-k, n - k, s);
}
int main(int argc, char *argv[])
{
uniqPerms(atoi(argv[1]));
return 0;
}
try this
A[n-1]=0;
func(n-1);
A[n-1]=1;
func(n-1);
//Think simple people but please bear with me i love java
//Assume array A is globally defined
void Binary(int n)
{
if(n<1)
{
System.out.println(A);
}
else
{
A[n-1]=0;
Binary(n-1);
A[n-1]=1;
Binary(n-1);
}
}
here is the recursive solution
#include <iostream>
#include <vector>
using namespace std;
char v[4];
int count = 0;
void printString(){
int i;
for(i = 0; i < 4; i++){
cout << v[i] << " ";
}
cout <<count << endl;
}
void binary(int n){
if(n < 0){
if(count == 2)
printString();
}
else{
v[n] = '0';
binary(n - 1);
v[n] = '1';
count++;
binary(n-1);
count--;
}
}
int main(){
binary(3);
return 0;
}
#include<stdio.h>
int main(){
int n,k,i,j,a[50];
//lets suppose maximum size is 50
printf("Enter the value for n");
scanf("%d",&n);
printf("Enter the value for k");
scanf("%d",&k);
//create an initial bitstring of k 1's and n-k 0's;
for(i=0;i<n;i++){
if(k>0)
a[i]=1;
else
a[i]=0;
k--;
}
for(i=0;i<n;i++){
if(a[i]==1){
for(j=0;j<n;j++){
if(j!=i&&a[j]==0){
a[j]=1;a[i]=0;
for(k=0;k<n;k++){printf("%d\n",a[k]);}
a[i]=1; a[j]=0;
}}}}
return 0;
}
**If Complexity doesn't matter you can use the following code which are done in java. which will provide the desired output in o(2^n).Here I have find all the combination of 0 and 1 for the given n bits in array of size n.In case of K bit is set I have counted the number of 1 presented is equal to k using countBits() funtion.if so I have printed that array.
public class GenerateAllStringOfNBitsWithKBitsSet {
public static int a[] ={0,0,0,0,0};
static int k=3;
public static boolean countBits(){
int y=0;
for(int i=0;i<a.length;i++)
y += a[i] & 1 ;
if(y==k)
return true;
return false;
}
public static void gen(int n)
{
if(n<1)
{
if(countBits())
System.out.println(Arrays.toString(a));
}
else
{
a[n-1]=0;
gen(n-1);
a[n-1]=1;
gen(n-1);
}
}
public static void main(String[] args) {
GenerateAllStringOfNBitsWithKBitsSet.gen(a.length);
}
}