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);
}
Related
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);
}
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
);
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.
Well it is a problem about finding the biggest and smallest number in a group of numbers, but we do not know how many numbers the user wants-
So far this is what i have done:
#include <stdio.h>
#include <conio.h>
int main()
{
int num;
int i;
int maxi=0;
int minim=0;
int cont = 0;
printf ("\nQuantity of numbers?: ");
scanf ("%d", &num);
while (num>0)
{
printf ("\nEnter number:");
scanf ("%d", &i);
if (num>i)
minim=i++;
else
if (i>num)
max=i++;
cont++;
}
printf ("\nBiggest number is es: %d", maxi);
printf ("\nSmallest number is: %d", minim);
getch();
return 0;
}
I did my program to ask how many numbers the user will want to put and i made the program to read them, BUT when it reads the biggest or/and smallest numbers it will sometimes changes biggest with small and it will not read negative numbers.
How do i do to make my program better?
You're comparing against the wrong values.
do
{
printf("Enter a number.\n");
scanf("%i", &input);
if min > input
min = input
if max < input
max = input
} while (input > 0);
#include <stdio.h>
#include <conio.h>
#include <limits.h>
int main(){
int num;
int i;
int maxi=0;
int minim=INT_MAX;
int cont = 0;
printf ("\nQuantity of numbers?: ");
scanf("%d", &num);
if(num > 0){
while (num>0){
printf ("\nEnter number:");
if(scanf("%d", &i) == 1 && !(i<0)){
if(minim > i)
minim = i;
if (maxi < i)
maxi = i;
++cont;
--num;
} else {
//fprintf(stderr, "redo input!\n")
;
}
scanf("%*[^\n]%*c");
}
printf ("\nBiggest number is : %d", maxi);
printf ("\nSmallest number is : %d\n", minim);
}
getch();
return 0;
}
You should initialize mini to the largest possible int, i.e. INT_MAX and maxi to the smallest possible int, i.e., INT_MIN. This way, even if the first number is negative, it will be considered for maxi, and if the first number is positive it will still be considered for mini. The constants INT_MAX and INT_MIN are included in <climits> or <limits.h>.
Also, you are comparing the current entered number with num, which is the counter of numbers entered by user, not one of the values he wants to compare. A better modified code would be :
#include<limits.h>
#include<stdio.h>
int main()
{
int num;
int maxi=INT_MIN; //initialize max value
int mini=INT_MAX; //initialize min value
int temp;
scanf("%d", &num); //take in number of numbers
while(num--) //loop "num" times, num decrements once each iteration of loop
{
scanf("%d", &temp); //Take in new number
if(temp>maxi) //see if it is new maximum
maxi=temp; //set to new maximum
if(temp<mini) //see if new minimum
mini=temp; //set to new minimum
}
printf("\nMaxi is:\t%d\nMini is:\t%d\n", maxi, mini); //print answer
return 0;
}
This is a program that gets numbers input. From the numbers given or inputted, store in an array those numbers only that are even. Input will stop/terminates once 5 even numbers are already stored in the array. So here's my code:
#include <stdio.h>
#include <conio.h>
int main()
{
int num[5];
int x, counter, even[5], numEven=0;
for(counter=0; counter<5; counter++){ //loop for getting the numbers from the user
printf("Enter number: ");
scanf("%d", &num[counter]);
if(num[counter]%2==0){ //storing the even numbers
even[numEven] = num[counter];
numEven++;
}
}
printf("\n\nEven numbers: "); //printing even numbers
for(counter=0; counter<numEven; counter++){
printf("%d, ", even[counter]);
}
getch();
return 0;
}
I have confusion in the part where will I stop the inputting when there's already 5 even numbers stored. Is there something missing? Or am I doing the wrong way? I hope I can get help and suggestions with the code. Thank you very much.
#include <stdio.h>
#include <conio.h>
int main()
{
int x, even[5], numEven = 0;
while (numEven < 5)
{
scanf("%d", &x);
if (x % 2 == 0)
{
even[numEven++] = x;
}
}
printf("\n\nEven numbers: "); //printing even numbers
for(x=0; x<numEven; x++)
{
printf("%d, ", even[x]);
}
getch();
return 0;
}
You keep readin inputs till numEven reaches 5. If the read input is an even number store it in the array and increment numEven.
Use a while loop if the number of times the program will ask the user for input is not fixed and dependent on the user's input.
while (numEven < 5) {
printf("Enter number: ");
scanf("%d", &num[counter]);
if (num[counter] % 2 == 0) {
even[numEven] = num[counter];
numEven++;
}
}