I made a program so when the user selects option 1 they can input a numerator and then a denominator, if they select option 2 it will print that fraction. However i want it to print the fraction in mixed form. For example if I did 20/3 it would print as 6 2/3. Any help as to how i would approach this problem?
This is my code ( hopefully easy to read (: )
#include <stdio.h>
#include <stdlib.h>
//Struct to hold fraction data
typedef struct fraction
{
int numerator, denom;
}fraction;
int main()
{
//Array of 100 fractions
fraction arrFraction[100];
int i = 0, j, num = 1;
//Loop till user want to stop
while (num == 1)
{
int choice;
printf("\nPress 1 to enter a fraction\n");
printf("Press 2 to view stored fractions\n");
scanf("%d", &choice);
if(choice == 1)
{
//Prompting user
printf("\nEnter your fraction, numerator followed by denominator\n ");
//Reading values from user
scanf("%d %d", &arrFraction[i].numerator, &arrFraction[i].denom);
//Incrementing counter
i++;
}
if (choice == 2) {
for (j = 0; j < i; j++)
{
//Printing fractions
printf("\n %d / %d \n", arrFraction[j].numerator, arrFraction[j].denom);
}
}
}//end of while loop
return(0);
}
You can use division(/) and modulus(%) operators.
printf("%d %d/%d",
arrFraction[j].numerator / arrFraction[j].denom ,
arrFraction[j].numerator / arrFraction[j].denom ,
arrFraction[j].denom
);
Related
i have a homework but i cant get the answer
I need to write a program in C...
Here is what is needed: You need to enter "n" natural number as input , and from all the natural numbers smaller than "n" , its needed to print the number which has the highest sum of devisors.
For exp: INPUT 10 , OUTPUT 8
Can anyone help me somehow?
I would really appreciate it !
i tried writing a program for finding the devisor of a number but i cant get far from here
#include <stdio.h>
int main() {
int x, i;
printf("\nInput an integer: ");
scanf("%d", &x);
printf("All the divisor of %d are: ", x);
for(i = 1; i < x; i++) {
if((x%i) == 0){
printf("\n%d", i);
}
}
}
I have implemented using function which will takes input number from user and then return the sum of divisor. hope this is one you looking for
/* function to return of sum of divisor
** input: x: integer number from user input
** return sum: sum of divisor of x
*/
int sum_of_divisor(int x)
{
int sum = 0;
for(int i = 1; i < x; i++)
{
if((x%i) == 0)
{
printf("%d\n", i);
sum = sum+i;
}
}
return sum;
}
int main() {
int x, i;
printf("\nInput an integer: ");
scanf("%d", &x);
printf("All the divisor of %d are: ", x);
printf("the sum of divisor is %d ", sum_of_divisor(x));
return 0;
}
Output:
Input an integer: 10
All the divisor of 10 are: 1
2
5
the sum of divisor is 8
After checking if i is a divisor of x, you should then store that value in another variable, for example m.
Repeat until a new divisor i is higher than that number. Add this new value to m.
disclaimer: I'm new to programming
I'm working on this problem
so far ive written this which takes user inputs and calculates an average based on them
#include <stdio.h>
int main()
{
int n, i;
float num[100], sum = 0.0, average;
for(i = 0; i < n; ++i)
{
printf("%d. Enter number: ", i+1);
scanf("%f", &num[i]);
sum += num[i];
}
average = sum / n;
printf("Average = %.2f", average);
return 0;
}
I'd like the user to enter -1 to indicate that they are done entering data; I can't figure out how to do that. so if possible can someone explain or give me an idea as to how to do it
Thank you!
#include <stdio.h>
int main()
{
int i = 0;
float num[100], sum = 0.0, average;
float x = 0.0;
while(1) {
printf("%d. Enter number: ", i+1);
scanf("%f", &x);
if(x == -1)
break;
num[i] = x;
sum += num[i];
i++;
}
average = sum / i;
printf("\n Average = %.2f", average);
return 0;
}
There is no need for the array num[] if you don't want the data to be used later.
Hope this will help.!!
You just need the average. No need to store all the entered numbers for that.
You just need the number inputs before the -1 stored in a variable, say count which is incremented upon each iteration of the loop and a variable like sum to hold the sum of all numbers entered so far.
In your program, you have not initialised n before using it. n has only garbage whose value in indeterminate.
You don't even need the average variable for that. You can just print out sum/count while printing the average.
Do
int count=0;
float num, sum = 0;
while(scanf("%f", &num)==1 && num!=-1)
{
count++;
sum += num;
}
to stop reading at -1.
There is no need to declare an array to store entered numbers. All you need is to check whether next entered number is equal to -1 and if not then to add it to the sum.
Pay attention to that according to the assignment the user has to enter integer numbers. The average can be calculated as an integer number or as a float number.
The program can look the following way
#include <stdio.h>
int main( void )
{
unsigned int n = 0;
unsigned long long int sum = 0;
printf("Enter a sequence of positive numbers (-1 - exit): ");
for (unsigned int num; scanf("%u", &num) == 1 && num != -1; )
{
++n;
sum += num;
}
if (n)
{
printf("\nAverage = %llu\n", sum / n);
}
else
{
puts("You did not eneter a number. Try next time.");
}
return 0;
}
The program output might look like
Enter a sequence of positive numbers (-1 - exit): 1 2 3 4 5 6 7 8 9 10 -1
Average = 5
If you need to calculate the average as a float number then just declare the variable sum as having the type double and use the corresponding format specifier in the printf statement to output the average.
I have to write a program in C that will take a base b from the user (assuming b is between 2 and 10), a natural number n and then n numbers that represent the digits of some number m in base b. The program should print out what decimal number m was input. For example, if you put b=5 and n=4 and then the numbers 3 ,4, 2 and 1 the program should output 486 because m=3*5^3+4*5^2+2*5^1+1*5^0=486
Note: You can assume that the digits will be the numbers between 0 and b-1.
So here's what I've done:
#include<stdio.h>
#include<math.h>
int main(void) {
int x,n,b,k=0,num=0,i,j;
scanf("%d", &b);
scanf("%d", &n);
for(i=1; i<=n; i++) {
scanf("%d", &x);
for(j=1; j<b; j++){
if(j>k){
num=num+x*(pow(b,n-j));
k=j;
break;
}
}
}
printf("m=%d", num);
return 0;
}
Can you tell me why this doesn't work for the numbers given in the example above? It outputs 485 instead of 486, while if I take for example b=7, n=3 and then numbers 5, 6 and 1, I get the correct solution m=288.
I suggest checking the return value of scanf(), Something like this is the right idea:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int
main(int argc, char *argv[]) {
int base, n, i, x, sum = 0, power;
printf("Enter base: ");
if (scanf("%d", &base) != 1) {
printf("Invalid base.\n");
exit(EXIT_FAILURE);
}
printf("Enter n: ");
if (scanf("%d", &n) != 1) {
printf("Invalid n.\n");
exit(EXIT_FAILURE);
}
power = n-1;
printf("Enter numbers: ");
for (i = 0; i < n; i++) {
if (scanf("%d", &x) != 1) {
printf("Invalid value.\n");
exit(EXIT_FAILURE);
}
sum += x * pow(base, power);
power--;
}
printf("Sum = %d\n", sum);
return 0;
}
Input:
Enter base: 5
Enter n: 4
Enter numbers: 3 4 2 1
Output:
Sum = 486
You need some small change to your logic.
#include <stdio.h>
#include <math.h>
int main(void) {
int x, n, b, num = 0, i;
scanf("%d", &b);
scanf("%d", &n);
for (i = 1; i <= n; i++) {
scanf("%d", &x);
num += x * pow(b, n - i);
}
printf("m=%d", num);
return 0;
}
Test
gcc -Wall main.c -lm
$ ./a.out
5
4
3
4
2
1
m=486
Test 2
./a.out
7
3
5
6
1
m=288
OK, so given a binary number, we can output a decimal number very easily. Just printf("%d%\n", x);
Next job is to convert a number given digits and base into a binary (machine representation) number.
int basetointeger(const char *digits, int b)
{
assert(b >= 2 && b <= 10);
// code here
return answer;
}
Now hook it all up to main
int main(void)
{
int x;
int base;
char digits[64]; // give more digits than we need, we're not worrying about oveflow yet
/* enter base *?
code here
/* enter digits */
code here
x = basetointger(digits, base);
printf("Number in decimal is %d\n, x);
}
For my program I am trying to ask the user to insert a fraction then if they hit option 2 it will display those fractions. So far I figured out how to ask the user for a number and it stores that number, how ever I need the code to ask to enter a numerator then when I hit enter I would enter the denominator. I believe my code to view the numbers is written correctly however I need it to display fractions, which is what I am having difficulty with.
This is my code:
#include <stdio.h>
#include <string.h>
#include <math.h>
int main()
{
int i = 0;
int num = 0;
while (num == 0) {
int num1;
printf("\nPress 1 to enter a the fraction\n");
printf("Press 2 view the fractions\n");
scanf("%d", &num1);
int num[100][500];
int n[100];
if (num1 == 1)
{
printf("Enter the numerator followed by the denominator\n");
scanf("%s", n);
strcpy(num[i], n);
i++;
}
if (num1 == 2)
{
printf("\n-----------------------------");
for (int j = 0; j < i; j++)
{
printf("\n%s\n", &num[j]);
}
printf("\n\n-----------------------------");
}
}
system("pause");
return(0);
}
I'm self-studying C and I'm trying to make 2 programs for exercise:
the first one takes a number and check if it is even or odd;
This is what I came up with for the first one:
#include <stdio.h>
int main(){
int n;
printf("Enter a number that you want to check: ");
scanf("%d",&n);
if((n%2)==0)
printf("%d is even.",n);
else
printf("%d is odd.",n);
return 0;
}
the second one should take n numbers as input and count the number of even numbers, odd numbers, and zeros among the numbers that were entered. The output should be the number of even numbers, odd numbers, and zeros.
I would like to ask how to implement the loop in this case: how can I set an EOF value if every integer is acceptable (and so I cannot, say, put 0 to end)? Can you show me how to efficiently build this short code?
#include <stdio.h>
int main(void) {
int n, nEven=0, nOdd=0, nZero=0;
for (;;) {
printf("\nEnter a number that you want to check: ");
//Pressing any non-numeric character will break;
if (scanf("%d", &n) != 1) break;
if (n == 0) {
nZero++;
}
else {
if (n % 2) {
nEven++;
}
else {
nOdd++;
}
}
}
printf("There were %d even, %d odd, and %d zero values.", nEven, nOdd, nZero);
return 0;
}
Check the return value of scanf()
1, 1 field was filled (n).
0, 0 fields filled, likely somehtlig like "abc" was entered for a number.
EOF, End-of-file encountered (or rarely IO error).
#include <stdio.h>
int main(void) {
int n;
for (;;) {
printf("Enter a number that you want to check: ");
if (scanf("%d",&n) != 1) break;
if((n%2)==0)
printf("%d is even.",n);
else
printf("%d is odd.",n);
}
return 0;
}
Or read the count of numbers to subsequently read:
int main(void) {
int n;
printf("Enter the count of numbers that you want to check: ");
if (scanf("%d",&n) != 1) Handle_Error();
while (n > 0) {
n--;
printf("Enter a number that you want to check: ");
int i;
if (scanf("%d",&i) != 1) break;
if((i%2)==0) {
if (i == 0) printf("%d is zero.\n",i);
else printf("%d is even and not 0.\n",i);
}
else
printf("%d is odd.\n",i);
}
return 0;
}
hey look at this
#include<stdio.h>
#include<conio.h>
void main()
{
int nodd,neven,num,digit ;
clrscr();
printf("Count number of odd and even digits in a given integer number ");
scanf("%d",&num);
nodd = neven =0; /* count of odd and even digits */
while (num> 0)
{
digit = num % 10; /* separate LS digit from number */
if (digit % 2 == 1)
nodd++;
else neven++;
num /= 10; /* remove LS digit from num */
}
printf("Odd digits : %d Even digits: %d\n", nodd, neven);
getch();
}
You can do something like this:
#include <stdio.h>
int main(){
int n,evenN=0,oddN=0,zeros=0;
char key;
do{
clrscr();
printf("Enter a number that you want to check: ");
scanf("%d",&n);
if(n==0){
printf("%d is zero.",n);
zeros++;
}
else if((n%2)==0){
printf("%d is even.",n);
evenN++;
}
else{
printf("%d is odd.",n);
oddN++;
}
puts("Press ENTER to enter another number. ESC to exit");
do{
key = getch();
}while(key!=13 || key!=27) //13 is the ascii code fore enter key, and 27 is for escape key
}while(key!=27)
clrscr();
printf("Total even numbers: %d",evenN);
printf("Total odd numbers: %d",oddN);
printf("Total odd numbers: %d",zeros);
return 0;
}
This program ask for a number, evaluate the number and then ask to continue for another number or exit.