Check for Fibonacci Prime in the amount of terms - c

I'm basically done with this, the only problem is when it loops to check for the prime number, it prints out prime numbers that are not in the Fibonacci sequence as well. Here is my code:
int main()
{
int no1,no2,newno,pno,i,terms,j;
no1 = 0;
no2 = 1;
printf("**Fibonacci and Prime Numbers**\n\n");
printf("Enter number of terms: \n");
scanf("%d", &terms);
printf("\nAmong the first %d terms of Fibonacci series that are also prime number: \n", terms);
for(i=0; i<terms; i++){
if(i<=no2){
newno = i;
}
else{
newno = no1+no2;
no1 = no2;
no2 = newno;
}
}
for(pno=2;pno<=newno;pno++){
for(j=2;j<=pno;j++){
if(pno%j==0){
break;
}
}
if(pno==j){
printf("%d \n", pno);
}
}
getch();
return 0;
}
I'm guessing its because of the pno++ , Am i right?

Try this
for(i=0; i<terms; i++){
newno = no1+no2;
no1 = no2;
no2 = newno;
for(j=2;j<=newno;j++){
if(newno%j==0){
break;
}
}
if(newno==j){
printf("%d prime\n", newno);
}
}

This is because you are checking only last number. You should check every number you generate. Consider creating function e.g. is_prime and check every fibonnaci number you compute. Result could look like this (taken from C - how to test easily if it is prime-number?)
int is_prime(int num)
{
if (num <= 1) return 0;
if (num % 2 == 0 && num > 2) return 0;
for(int i = 3; i < num / 2; i+= 2)
{
if (num % i == 0)
return 0;
}
return 1;
}
And your main loop that calls is_prime for every number:
for (i = 0; i<terms; i++) {
if (i <= no2) {
newno = i;
}
else {
newno = no1 + no2;
no1 = no2;
no2 = newno;
if (is_prime(newno)) {
printf("%d\n", newno);
}
}
}

This is just a rough try of what you are trying to achieve .
#include<stdio.h>
int main()
{
int no1,no2,newno,pno,i,terms,j;
no1 = 0;
no2 = 1;
printf("**Fibonacci and Prime Numbers**\n\n");
printf("Enter number of terms: \n");
scanf("%d", &terms);
printf("\nAmong the first %d terms of Fibonacci series that are also prime number: \n", terms);
for(i=0; i<terms; i++)
{
newno = no1+no2;
no1 = no2;
no2 = newno;
for(pno = 2; pno <= newno ;pno++)
{
if(pno == newno)
printf("%d\n",newno);
if(newno%pno == 0)
break;
}
}
return 0;
}
You were only trying to check only the last term and that to in incorrect way.
The for loop you used
for(pno=2;pno<=newno;pno++){
for(j=2;j<=pno;j++){
if(pno%j==0){
break;
}
}
if(pno==j){
printf("%d \n", pno);
}
}
was applying check on local counter pno that does not make any sense.
You can further optimize your program using a better way to find if a number is prime as it will take huge time for large inputs.

Related

Why does my code calling a function twice?

#include <stdio.h>
int bolucu(int n){
int temp;
temp=n;
int basamak=0;
while(temp != 0){
temp/=10;
++basamak;
}
int digits = 0;
int m = n;
while (m) {
digits++;
m /= 10;
}
digits /= 2;
int tmp = 0, lower_half = 0;
while (digits--) {
tmp *= 10;
tmp += n % 10;
n /= 10;
}
while (tmp) {
lower_half *= 10;
lower_half += tmp % 10;
tmp /= 10;
}
if (basamak % 2==1){
n/=10;
}
int a;
int b;
a = n;
b=lower_half;
printf("%d %d\n",a,b);
int loopTemp;
for(int i=0;i<10;i++){
a=3*a+2;
b=2*b+3;
if(a>b){
temp=a;
a=b;
b=temp;
}
if(a==b){
printf("Congratulations you caught one!!!\n");
return 1;
break;
}
}
if(a!=b){
printf("10 tries were not enough!\n");
return 2;
}
}
int main()
{
int number;
printf("\nEnter a number with at least two digits: ");
scanf("%d",&number);
bolucu(number);
while(bolucu(number) != 1){
printf("\nEnter a new number: ");
scanf("%d",&number);
printf("%d",bolucu(number));
}
return 0;
}
e.g:
This is terminal screen.
As you can see there is a second one. First one is true but i don't want second one.
How can i get rid of the second calling?
(Also sorry for bad code writing, i'm new)
What im missing here?
And i cant use any library other than stdio.(Like math.h)
The reason for this is because you call the bolucu() function both inside the while loop and in it's condition check. To fix this, call the function and hold it's result in a variable once, and then use that single result in both the check and your print statement. Your main function can be rewritten like so:
int main()
{
int number;
printf("\nEnter a number with at least two digits: ");
scanf("%d", &number);
int result = bolucu(number);
while (result != 1)
{
printf("\nEnter a new number: ");
scanf("%d", &number);
result = bolucu(number);
printf("%d", result);
}
return 0;
}

C Program (Prime Number in a given range)

I have started learning C language. I wrote this program to find all prime numbers between the given range but I am unable to get the expected output.
Can anyone tell me what's wrong with this program please?
#include <stdio.h>
#include <conio.h>
void main() {
int min, max, i, j, count = 0;
printf("Enter Your First Number\n");
scanf("%d", &min);
printf("Enter Your Last Number\n");
scanf("%d", &max);
for(i=min; i<=max; i++) {
for(j=1; j<=i; j++) {
if(i % j == 0) {
count++;
}
}
if(count==2) {
printf("%d\t",i);
}
}
getch();
}
I just suggest getting rid of that count variable.
How do you know if a number N is prime? If for every j in the range (2 to N-1) you have N%j != 0.
So:
In the inner loop, use j from 2 to N-1 (instead of from 1 to N as you used tio do). In fact N%1 and N%N will be 0
The first time you find a j so that N % j == 0 break. You are sure it's not prime
Why incrementing count? For a prime number the j counter will be equal to i (because you looped until j<i, and the last j++ made j
equal to i). So just check for j == i and print the prime number i
#include <stdio.h>
#include <conio.h>
int main( void )
{
int min, max, i, j, count = 0;
printf("Enter Your First Number\n");
scanf("%d", &min);
printf("Enter Your Last Number\n");
scanf("%d", &max);
for(i=min; i<=max; i++)
{
// Was for(j=1; j<=i; j++)
for(j=2; j<i; j++)
{
if(i % j == 0)
{
//Was count++;
break;
}
}
//Was if(count==2)
if(j == i)
{
printf("%d\t",i);
}
}
getch();
return 0;
}
Here you are.
#include <stdio.h>
int main( void )
{
printf( "Enter the range of numbers (two unsigned integer numbers): " );
unsigned int first = 0, last = 0;
scanf( "%u %u", &first, &last );
if ( last < first )
{
unsigned int tmp = first;
first = last;
last = tmp;
}
do
{
int prime = first % 2 == 0 ? first == 2 : first != 1;
for ( unsigned int i = 3; prime && i <= first / i; i += 2 )
{
prime = first % i != 0;
}
if ( prime ) printf( "%u ", first );
} while ( first++ != last );
putchar( '\n' );
return 0;
}
The program output might look like
Enter the range of numbers (two unsigned integer numbers): 0 100
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
As for your program then you need re-initialize the variable count before the inner loop
for(i=min; i<=max; i++) {
count = 0;
for(j=1; j<=i; j++) {
if(i % j == 0) {
count++;
}
}
And the inner loop is inefficient.
Need to reset the value of count. It starts at count=0, then for any inputs, the loops will count up. The For each outer loop index, it will go like this:
1 (1%1=0 --> count++, count = 1)
2 (2%1=0 --> count++, and 2%2=0 --> count++, count = 3)
3 (3%1=0 --> count++, and 3%3=0 --> count++, count = 5)
etc... until max is reached.
You can use a simple isprime function to check whether a number is prime or not and then call the function for the given interval.
To find whether a number is prime or not , we can use a simple primality test to check it.
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
bool isprime(int n)
{
if(n <= 1) return false;
if(n <= 3) return true;
if(n%2 == 0 || n%3 == 0) return false;
for(int i = 5;i*i <= n;i += 6)
{
if(n%i == 0 || n%(i + 2) == 0)
{
return false;
}
}
return true;
}
int main()
{
int a,b;
printf("Enter the first number :");
scanf("%d",&a);
printf("Enter the second number :");
scanf("%d",&b);
for(int i = a;i <= b;i++)
{
if(isprime(i)) printf("%d ",i);
}
return 0;
}
There is a simple change you should do:
#include <stdio.h>
#include <conio.h>
void main() {
int min, max, i, j, count;
printf("Enter Your First Number\n");
scanf("%d", &min);
printf("Enter Your Last Number\n");
scanf("%d", &max);
for(i=min; i<=max; i++)
{
count=1;
for(j=2; j<=i; j++)
{
if(i % j == 0) {
count++;
}
}
if(count==2) {
printf("%d\t",i);
}
}
}
My answer may be a bit late, but since it's the same issue, i'll write it here in case it helps someone else coming to this thread in the future.
My code is written from the POV of a beginner (No complex functions or data types are used) as this is a code that mostly they will get stuck on.
Working:
User inputs the range.
Using the for loop, each number in the range is sent to the isprime function which returns TRUE or FALSE after checking the condition for being a prime number.
if TRUE : program prints the number.
if FALSE : program skips the number using continue function.
#include<stdio.h>
int isprime(int num);
int main() {
int min, max;
printf("Input the low number: ");
scanf("%d", &min);
printf("Input the high number: ");
scanf("%d", &max);
for(int i = min; i<=max; i++) {
if(isprime(i) == 1) {
printf("%d ", i);
}
else if(isprime(i) == 0){
continue;
}
}
return 0;
}
int isprime(int num) {
int count = 0;
for(int i=2; i<=(num/2); i++) {
if(num % i == 0 ) {
count ++;
}
else{
continue;
}
}
if(count>0){
return 0;
}
else if (count == 0){
return 1;
}
}

no repeating values in array

I need to make a program that stores numbers inside of an array. But it must have no duplicate elements.
int x;
int z[8];
for( x = 0; x<8;x++)
printf("number: ");
scanf("%d",&z[x]);
}
for( x=0;x<8;x++) {
printf("%d ",z[x]);
}
First, initialize the array, so that you do not end up reading an uninitialized value and fail the test.
int user_nums[6] = {0};
Next, you need to have another check in the for loop, to read the number again if it is a duplicate.
The code will look like this.
#include<stdio.h>
int main(){
int x,y;
int exists = 0;
int user_nums[6] = {0};
for( x = 0; x<6;x++){//for loop to get the players selected numbers
do {
exists = 0;
printf("Enter a number(from the #'s 1-42): ");
scanf("%d",&user_nums[x]);
for(y =0; y < x; y++) { //to check for duplicates
if (user_nums[x] == user_nums[y])
{
printf("Number already exists\n ");
exists = 1;
break;
}
}
}while (user_nums[x]<1 || user_nums[x]>42 || exists);//accepts only numbers from 1-42 which are not duplicates (continous to ask you for a number until condition is met).
}
printf("Your numbers: \n");
for( x=0;x<6;x++){
printf("%d ",user_nums[x]); // prints the numbers you inputed.
}
return 0;
}
The following code could work in O(n):
#include<stdio.h>
int main()
{
int user_nums[6];
int index[50];
for (int i = 0; i != sizeof(index) / sizeof(index[0]); ++i)
index[i] = -1;
for (int i = 0; i < sizeof(user_nums) / sizeof(user_nums[0]); ++i) {
for (;;) {
printf("Enter a number(from the #'s 1-42): ");
scanf("%d", user_nums + i);
if (user_nums[i] < 1 || user_nums[i] > 42) {
printf("wrong number\n");
continue;
}
if (index[user_nums[i]] != -1) {
printf("dump number\n");
continue;
}
index[user_nums[i]] = i;
break;
}
}
printf("Your numbers: \n");
for(int i = 0; i < 6; ++i)
printf("%d ", user_nums[i]);
return 0;
}

C programming - do while loop help (code almost done)

I have created a code that writes out the sum of all even numbers.
But every time it loops the sum of the last run is saved and added in the sum of the new run. how do i make it have a new loop?
sorry for my bad english, and thx in advance
int main()
{
int number = 0;
int sum = 0;
printf("Welcome to\"Sum Evens\"!");
do
{
printf("\ninput a number: ");
scanf(" %d", &number);
if (number == 0)
{
printf("Goodbye, have a nice day!\n");
break;
}
printf("\nSum:");
if (number % 2 != 0)
{
number -= 1;
}
for (int i = 0; i <= number; i += 2)
{
printf(" %d ", i);
if (i != number)
{
printf("+");
}
sum += i;
}
printf("= %d\n", sum);
} while (number != 0);
system("pause");
return 0;
}
Watch these sort of problems melt away if you declare your variables as close to their first use as possible.
In your case, move int sum = 0; to just before the for loop.

Detect if the number in the array has been already input (C program)

I am kind of confused. I'd like to make a program where if the number in the array has been already input, then it will detect it and say it was repeated, so the program would tell the user to put another non-repeated integer.
#include <stdio.h>
#define SIZE 5
int main()
{
int array[SIZE];
int i;
int j;
for (i = 0; i < SIZE; i++)
{
printf("[%d] Insert a number: ", i + 1);
scanf("%d", &array[i]);
j = i - 1; // This is the closest that I've gotten guys. But I need to create a loop to make j be -1 until it finds a repeated number in the array.
if (array[i] == array[j])
{
printf("The number is repeated");
i--;
}
if (array[i] > 1000)
{
printf("Sorry, the number you entered cannot be bigger than 1000\n");
i--;
}
if (array[i] < 0)
{
printf("Sorry, the number you entered cannot be less than 0\n");
i--;
}
}
for (i = 0; i < SIZE; i++)
{
printf("The array inside is %d\n", array[i]);
}
return 0;
}
As you can see, I did something similar. I just put j = i - 1 so basically it will tell the program that it was repeated. However, I suppose that I should create a loop that will subtract -1 to j until it finds the repeated value (if there is one). I just not have any idea how to create that loop and make it work.
Thank you very much!
The checks can be done the following way (without testing)
int array[SIZE];
int i;
for (i = 0; i < SIZE; i++)
{
int valid = 1;
int num;
do
{
printf("[%d] Insert a number: ", i + 1);
scanf("%d", &num );
if ( !( valid = !( num > 1000 ) ) )
{
printf("Sorry, the number you entered cannot be bigger than 1000\n");
}
else if ( !( valid = !( num < 0 ) ) )
{
printf("Sorry, the number you entered cannot be less than 0\n");
}
else
{
int j = 0;
while ( j < i && num != array[j] ) j++;
if ( !( valid = j == i ) )
{
printf("The number is repeated");
}
}
} while ( !valid );
array[i] = num;
}
This should work for you:
#include <stdio.h>
#define SIZE 5
int main() {
int array[SIZE];
int numberCount, repeatCount;
for(numberCount = 0; numberCount < SIZE; numberCount++) {
printf("[%d] Insert a number:\n>", numberCount + 1);
scanf("%d", &array[numberCount]);
for(repeatCount = 0; repeatCount < numberCount; repeatCount++) {
if (array[numberCount] == array[repeatCount]) {
printf("\nThe numbe is repeated\n");
numberCount--;
break;
}
}
if (array[numberCount] < 0) {
printf("\nSorry, the number you entered cannot be less than 0\n");
numberCount--;
}
if (array[numberCount] > 1000) {
printf("\nSorry, the number you entered cannot be bigger than 1000\n");
numberCount--;
}
}
printf("\n\n");
for(numberCount = 0; numberCount < SIZE; numberCount++)
printf("The array inside is %d\n", array[numberCount]);
return 0;
}

Resources