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Calculating e^x but I'm really new to programming. I've made like two programs before. I've fixed any errors the shell points out yet get zero once it is run.
I have this :
#include <stdio.h>
int main (void) {
float answer = 1.0, x, next_term, n;
int power;
printf("Enter x: ");
scanf("%f", &x);
printf("Enter the number of terms: ");
scanf("%f", &n);
next_term = x;
for (power = 1; power <= n ; power = power + 1) {
answer = answer * next_term;
next_term = next_term * x / (power + 1);
}
printf("exp(%f) = %f\n", x, answer);
return 0;
}
Output:
exp(8.000000) = 0.000000
If you're calculating an infinite series, you want to add the terms, not multiply them... i.e. answer=answer+next_term;
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So i want my code actualy had different sum each cases, but the sum keep adding from the other cases, like for example the cases 2 sum are sum from loop 2nd + cases 1, the cases 4 sum are sum from cases 1,2,3 and loop number 4
the Output.
`
#include <stdio.h>
int main() {
int cases;
int day;
int animal;
int n;
int a;
int sum = 0;
animal = 0;
printf("Enter cases \n ");
scanf(" %d", &n);
for (cases = 0; cases < n; cases++)
{
printf("cases #%d\n", cases+1);
printf("Enter how many days.\n");
scanf(" %d", &a);
for(day=1;day<=a;){
printf("Enter how many animal that you capture at day #%d\n", day);
scanf(" %d", &animal);
day++;
sum = sum + animal;
}
animal = 0;
day = 1;
printf("cases#%d = %d\n", cases + 1, sum);
}
return 0;
}`
You need to return sum to 0 in every case
Put sum=0; under for (cases...
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This is the task
Input: 1 1
Output: 2
Input: -1 -1
Output: -2
Input: 1.1 2.2
Output: 3.3
I know how to do 2 of 3 task, but I don't know how to include float numbers when I have int numbers.
You can use %g to get expected outputs that you have mentioned.
#include <stdio.h>
int main()
{
float num1 = 0;
float num1 = 0;
float tot = 0;
printf("Enter number 1 = ");
scanf("%f", &num1);
printf("Enter number 2 = ");
scanf("%f", &num2);
tot = num1 + num2;
printf("Output = %g", tot);
return 0;
}
input : 2, 2
output :4
input : 5.5, 5
output: 10.5
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For an assignment I need to convert a 16-bit decimal number to a binary number. So for example the number 9 should print 0000000000001001. My professor started us with this code:
void printBinary(short n)
{
}
int main(int argc, char **argv)
{
short n;
printf("Enter number: ");
scanf("%hd", &n);
printBinary(n);
}
I am very confused as to where to go from here. I really would appreciate anyone helping me understand what to do, as I am very new to coding. Thanks in advance.
Try this:
int printBinary(int integer)
{
int remainder, counter = 0, my_binary = 0;
while (integer > 0)
{
remainder = integer % 2;
my_binary += remainder * pow(10, counter);
integer /= 2;
counter++;
}
printf ("Binary is %d\n", my_integer");
}
This has been taken from here.
NOTE: I have the assumption that you want to convert only positive decimals.
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I need to create a program that gives the sum of the all the numbers between the constants of a and b given by the user. b needs to be greater than a.
#include <stdio.h>
void main()
{
int index, begno, endno, sum = 0;
printf("Program for sum of all numbers in the given range\n");
printf("Enter Beg. No.: ");
scanf("%d", &begno);
printf("Enter End. No.: ");
scanf("%d", &endno);
index = begno;
for(; index <= endno; index ++)
sum = sum + index;
printf("The sum of even numbers between %d and %d is: %d", begno, endno, sum);
}
The code given looks OK, but if you want the sum without including the last number, as is generally the case you should change the for loop like this
for(; index < endno; index ++)
I would start by implementing a loop to compute:
$$\sum_{n=a}^{b}
n$$
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I'm a beginner and I am having a really hard time while doing this program.
The question is:
(1/1!)+(2/2!)+(3/3!)+(4/4!)- - - -n
So here are the n number of terms(in which a number is divided by its factorial) and I have to display the output of the sum of any number of terms which are given in scanf function.
Only one thing I know is that this program can be done by using "Nested for" loop but I haven't perfect grip yet on C language. So you guys have to have help me out in this. :)
#include <stdio.h>
#include <conio.h>
void main(void){
int s,a,b,n,fact=1;
//clrscr();
printf("Enter number of terms=");
scanf("%d",&n);
for(a=1;a<=n;a++) {
fact=fact*a;
b=(a/fact);
printf("Sum=%d",s);
}
getche();
}
P.S It's must for me to do it with "Nested for" loop.
No you do not need any Nested for loops to solve your problem. Here's a procedure you may follow:
function factorial
Input: numbers L.
Output: factorial of L.
function sum
Input: n.
Output: sum.
sum = 0;
for i = 1 to n, do
sum ← sum + (i / factorial(i))
return sum
#include <stdio.h>
int main(void) {
// your code goes here
int n;
float sum = 0,d,fact =1,j,i;
printf("Enter the number:");
scanf("%d",&n);
for(i=1;i<=n;i++){
fact = 1;
for (j = 1; j <= i; j++){
fact = fact * j;
}
d = (float) i / (float) fact ;
sum = sum + d;
}
printf("sum = %f", sum);
return 0;
}
Its working ..you can check over here :-https://ideone.com/JVXQVX