I have to write a program which calculates and displays the following expression:
P=(1/2-3/4)*(5/6-7/8)*...*[n/(n-1) - (n+2)/(n+3)]
I did this and I ran it and it shows no errors. When I run it, for every value that I enter, it shows P=0. What is wrong?
#include <stdio.h>
int main (void)
{
float s, p=1.0;
int i, n;
printf("Enter a number:");
scanf("%d", &n);
for (i=1;i<=n;++i) {
p*=((i)/(i+1)-(i+2)/(i+3));
}
printf("p=%f\n", p);
}
You're using integer arithmetic rather than floating point, so the division operations are being truncated.
Change:
p*=((i)/(i+1)-(i+2)/(i+3));
to:
p*=((float)(i)/(i+1)-(float)(i+2)/(i+3));
It is not working because i is integer: when you divide (i)/(i+1) or (i+2)/(i+3), you get a zero in both cases, because when i is positive, denominator is greater than numerator. Zero minus zero is zero, p times zero is zero, so that's what you get.
To fix this problem, declare i as float:
#include <stdio.h>
int main ()
{
float s,p=1.0, i;
int n;
printf("Put a number:");
scanf("%d",&n);
for (i=1;i<=n;++i) {
p*=((i)/(i+1)-(i+2)/(i+3));
}
printf("\n p=%f",p);
return 0;
}
You are using int variables, not float variables, thus i/(i+1) and (i+2)/(i+3) will always evaluate to 0 since it's integer division.
Just use some casts to force the evaluation using floating point arithmetics:
((float)i)/(i+1)-((float)i+2)/(i+3)
Your problem is that integer division truncates rather than returning the floating point results you appear to be looking for:
i/(i+1)
and
(i+2)/(i+3)
are always 0 in C.
Related
I'm trying to build a calculator, which input a number n it can automatically output the sum of
1/2+2/2 +1/3+2/3+3/3 +...+1/n+2/n+...+n/n and for each denominator, print the result once.
The calculator will restart and ask for input after output the final result of 1/2+2/2 +1/3+2/3+3/3 +...+1/n+2/n+...+n/n .And when the the user decides to stop the program enter 0.
for input=3 ,the output is sum =1 ,sum=2 ,yet I expected it to be sum=3/2, sum=7/2
Here's the program
#include <stdio.h>
int main(){
printf("Please enter the max denominator：\n");
int i,j,n;
int sum=0;
do{
scanf("%d",&n);
for (j=2;j<=n;j++){
for(i=1;i<=j;i++){
sum+= i/j;
if (i==j){
printf("sum=%d\n",sum);
break;
}
}
}
} while (n!=0);
return 0;
}
EDIT: revised part due to advice
int i,j,n;
double sum=0.0;
do{
scanf("%d",&n);
sum=0.0;
for (j=2;j<=n;j++){
for(i=1;i<=j;i++)
sum+=(double)i/j;
printf("sum=%lf\n",sum);
}
} while (n!=0);
return 0;
}
Your program is working with integers. i, j, n, and sum are all integers. This means they cannot encode numbers like 3/2 or 7/2.
So what happens when you divide two integers, like 1/2? The answer is that the result rounds down to the nearest integer. Since 1/2 is 0.5, the nearest integer below that is 0, so the result is 0.
You need to use float, which uses floating point arithmetic. Floating point arithmetic has its own set of fun rounding issues (You may be surprised to find that "0.2" cannot be represented exactly with floating point numbers).
A more exact alternative is to write your own functions to handle rational numbers as a pair of integers -- a numerator and a denominator. It will require additional algorithms (such as a Greatest Common Divisor algorithm), but it can describe these sorts of problems exactly.
#include<stdio.h>
int main()
{
int n,i;
float sum=0;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
printf("1/%d +",i);
sum=sum+1/i;
}
printf("= %f",sum);
return 0;
}
In this code I've kept the sum variable float to see the values after decimal. But the result doesn't appear to be correct. 000000 is shown after decimal. Why is this happening? What should I do to get the correct value without changing other variable's datatype? When I change other variables to float datatype,the answer remains correct.
Because both 1 and i are integers, the expression 1/i is evaluated as integer.
Therefore, if i > 1 then 1/i == 0.
You can solve this using float(1)/i or 1/float(i) or float(1)/float(i).
This question already has answers here:
Why dividing two integers doesn't get a float? [duplicate]
(7 answers)
Closed 5 years ago.
input example : 356
356/100, is suppused to be 3.56
But I'm getting 3.0000000000, I'm using ideone online compiler for C.
#include <stdio.h>
#include <math.h>
int main() {
int n;
double frac;
scanf("%d", &n);
frac = (n/100);
printf("%lf", frac);
return 0;
}
That's because here frac = (n/100); you are doing plain integer arithmetic (as n is declared as an int and 100 is interpreted as an int (any whole number is taken to be an int unless specified otherwise)). What you need to do is say explicitly that you want to do an arithmetic operation with digits after decimal point. One way is to use a cast: frac = ((double) n/100);
If you don't use the cast, the division will be performed as you expect, but then the digits after the decimal point will be dropped. Since frac is declared as a double, 0s would get tacked on to the end.
In C, the result of division of two integer numbers (e.g. int, short, long) is also an integer (it is counter-intuitive, but it is implemented this way for performance reasons). As a result, the result of 5/2 is 2 and not 2.5. This rule is only for integer numbers. So, if you need to get a floating-point result, at least one of the numbers in a division operation must be of a floating-point type.
In case of your code, if you use 100.0 instead of 100, you will get the desired result. Also you can use casts or define n as double.
This should work:
#include <stdio.h>
#include <math.h>
int main() {
int n; // You can define n as double but don't forget to modify scanf accordingly.
double frac;
scanf("%d", &n);
frac = (n/((double)100)); // Or, frac = (n/100.0)
printf("%lf", frac);
return 0;
}
You cannot call division using integers to be double without declaring it.
For example
int / int will result int.
Try declaring n as double
double n;
scanf("%lf", &n);
frag = n/100;
input data type is an integer.
just change it to double or float to fix this problem.
int main() {
double n;
double frac;
scanf("%d", &n);
frac = (n/100);
printf("%lf", frac);
return 0;
}
So I am trying to write a program which takes a floating point value and represents it in terms of a fraction. I am facing a weird problem. This is my program:
#include<stdio.h>
#include<math.h>
int gcd(int,int);
void main()
{
int n,n1,n2,g;
float a,a1;
printf("Enter number of digits after decimal point: ");
scanf("%d",&n);
printf("Enter number: ");
scanf("%f",&a);
n2=pow(10,n);
a1=a*n2;
n1=(int)a1;
g=gcd(n1,n2);
n1/=g;n2/=g;
printf("The number is %d/%d",n1,n2);
}
int gcd(int a,int b)
{
int x,flag=0;
int n1=((a>b)?a:b);
int n2=((a<b)?a:b);
for(x=n1;x>=1;x--)
{
if(n1%x==0 && n2%x==0)
{
flag=1;break;
}
}
if(flag==1)
return x;
else
return 1;
}
Now this program gives the correct answer for numbers with only 1 decimal point. For example:
Enter number of digits after decimal point: 1
Enter number: 2.5
The number is 5/2
However for number with two or more digits after decimal point it gives a wrong answer. For example:
Enter number of digits after decimal point: 2
Enter number: 2.50
The number is 247/99
I know that floating point numbers are not accurate but I did not expect this big a variation. Is there any way to work around this and make this program work???
Works for me. I believe the reason is that you use pow(10, n) and pow is inexact on your platform. Use a simple for-loop instead:
n2 = 1;
for (int c = 0; c < n; c++) {
n2 *= 10;
}
As functions like pow(10, n) may generate results near an integer value instead of an exact integer value. Rather than truncating with (int), use one of the round functions should you want to continue using pow().
Consider long int lround(double x), long int lroundf(float x) which round and covert to an integer type in one call.
The lround and llround functions round their argument to the nearest integer value, rounding halfway cases away from zero, regardless of the current rounding direction. C11 §7.12.9.7 2
n2 = lround(pow(10,n));
a1 = a*n2;
n1 = lroundf(a1);
This is a program to find number of digits.
Here is my code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main() {
int i = 0, n;
printf("Enter number: ");
scanf("%d", &n);
while ((n / pow(10, i)) != 0) {
i++;
}
printf("%d", i);
}
This program gives 309 as the output (value of i) on any input. However, if I store the value of pow(10, i) in another variable and put it in while loop, I get the correct output. Please help!
C++ uses the most precise type (when types are mixed) when doing a calculation or evaluation, and here you are effectively mixing a double with an int. You are dividing a user input number by a very large exponential number.
This in theory will never be zero, no matter how big the divisor gets. So, the theoretical result should be infinite. However, due to the limits of a double (which is returned by pow), it will still eventually approximate zero.
If you store the return of pow in an integer, you will no longer be mixing types and will effectively be working out the number of digits or what the approximate log base 10 is (any integer divide by a larger integer will equal zero).
Change:
while( ( n / pow(10,i) )!=0 ){
To:
while( ( n / pow(10,i) ) >= 1 ){
pow returns double and therefore the full result is double and the fractional part will be always non-zero until running out of exponent possible values which is 309 (translated to decimal, binary is 1024).