How i get the float result of the division?
Although i defined the average array as float.
int main()
{
const int Number = 20;
int Fibonacci[Number];
float average[Number];
for ( int i =0; i <= Number; i++ )
{
if ( i == 0 )
{Fibonacci[i] = 0;}
else if ( i == 1 )
{Fibonacci[i] = 1;}
else
{Fibonacci[i] = Fibonacci[i -1] + Fibonacci[i -2];
//average[i] = (Fibonacci[i -1] + Fibonacci[i -2])/2 ;
}
}
cout<< "The first 20 Fibonacci series numbers are: \n";
for ( int i = 1; i <= Number; i++)
{ cout<< Fibonacci[i]<<endl;
}
cout<< "The average adjacent array numbers are: \n";
for ( int i = 3; i <= Number; i++)
{ average[i] = (Fibonacci[i]/2);
//cout.precision(0);
Here the proplem
cout<< average[i]<<endl; <-----here the problem!!
}
return 0;
}
I appreciate any help.
Thanks in advance.
When you do the division, you are doing an integer division, so you won't get floating point results. A simple fix would be the following:
average[i] = Fibonacci[i] / 2.0f;
Note that if one of the operands to / is a float, then you get floating point division.
Also, note that your loops index too far into the array. You need to stop before Number, like this:
for ( int i = 0; i < Number; i++)
This is because valid array indexes go from 0 .. (Number - 1).
If Fibonacci[i] is of type int, then (Fibonacci[i]/2) is an integral division, resulting in an integral value (without any fractional part). Assigning this integral result to a float does not change the fact that you have performed an integral division.
You can enforce a floating point division by making one of the operands a floating point value.
So (1) use either a cast...
((float)Fibonacci[i])/2
or (2) divide by 2.0f (which is a float value):
Fibonacci[i]/2.0f
Just use typecasting the fibonacci array like this-
average[i] = (float)Fibonacci[i]/2;
Because in order to get float result any of the two variable used in / operation has to be float.
Related
In the C++ function below, why is numbers[max_index1] cast to long long and then multiplied by numbers[max_index2]? I would of thought that you'd multiply the numbers and then cast?
Also would it not make more sense to make the vector numbers type long long instead of int therefore the casting wouldn't be necessary?
long long MaxPairwiseProductFast(const vector<int>& numbers) {
int n = numbers.size();
int max_index1 = -1;
cout << "value at max_index1 is " << numbers[max_index1] << std::endl;
for(int i = 0; i < n; i++)
if((max_index1 == -1) || (numbers[i] > numbers[max_index1]))
max_index1 = i;
int max_index2 = -1;
for(int j = 0; j < n; j++)
if((numbers[j] != numbers[max_index1]) && ((max_index2 == -1) || (numbers[j] > numbers[max_index2])))
max_index2 = j;
return ((long long)(numbers[max_index1])) * numbers[max_index2];
}
int main() {
int n;
cin >> n;
vector<int> numbers(n);
for (int i = 0; i < n; ++i) {
cin >> numbers[i];
}
long long result = MaxPairwiseProductFast(numbers);
cout << result << "\n";
return 0;
}
((long long)(numbers[max_index1])) * numbers[max_index2];
numbers[max_index2] will be promoted to long long before multiplication is performed.
If you multiply two int's and the result overflowed, there is nothing you can achieve by casting that result to long long, so you cast first, then multiply.
Also would be not make more sense to make the vector numbers type long long instead of int therefore the casting wouldn't be necessary?
If you know that the individual numbers will fit an int, but the result of multiplying two int's can overflow, this will help save space.
I woudld of thought that you'd multiply the numbers and then cast?
Imagine your two operands have the value std::numeric_limits<int>::max(). This is the biggest value an int can represent, and (since it's a positive integer) the result of squaring this number is even bigger.
When you multiply two int values, the result is also an int. See here (specifically conversions, integral promotion and overflow of signed types).
Since the result is by definition bigger than the largest value you can store in an int, doing this multiplication with ints gives an undefined result. You need to perform the multiplication with a type large enough to store the result.
I want to find the sum up to the 'n'th term for the following series:
(1/2)+((1*3)/(2*4))+((1*3*5)/(2*4*6))....
So, I wrote the following program in c++ :
#include <bits/stdc++.h>
#include <conio.h>
using namespace std;
int main()
{
int p=1, k=1, n=0;
float h=0;
cout<<"Enter the term: ";
cin>>n;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=i; j++)
{
p*=((2*j)-1);
k*=(2*j);
}
h+=(p/k);
p=1;
k=1;
}
cout<<"The sum is : "<<h;
return 0;
getch();
}
However, the output of the program always gives me '0'. I can't figure out the problem with the program.
N.B. I'm new to programming.
The problem here is that you haven't declared p and k as float or doubleor explicitly cast them as such before the calculation and assignment to h.
What's happening is for every iteration of the loop p < k (by nature of the problem) since p and k are both declared as int, p / k = 0. So you're just summing 0 for every iteration.
Either declare p and k as float or double or do this:
h += ((float) p) / ((float) k)
Also, for this specific problem I assume you're looking for precision, so be wary and look into that as well Should I use double or float?
implicit conversion and type casting are a trap where all newbies fall.
in the instruction:
h += p/k;
the compiler performs an integer division first, then a promotion of the result to floating point type.
and since:
p < k ; for all i,j < n
then:
res = (p / k) < 1 => truncates to 0; // by integer division
thus:
sum(1->n) of p/k = sum (1->n) 0 = 0;
finally:
h = conversion to float of (0) = 0.0f;
that's why you have the result of 0.0f at the end.
the solution:
1- first of all you need to use the natural type for floating point of c++ which is "double" (under the hood c++ promotes float to double, so use it directly).
2- declare all your variable as double, except the number of terms n:
3- the number of terms is never negative, you need to express that in your code by declaring it as an unsigned int.
4- if you do step 3, make sure to catch overflow errors, that is if the user enters a negative number your risk to have a very big number in "n", expel : n =-1 converts to 0xffffffff positive number.
5- engineer your code sometimes is better.
6- include only the headers that you need, and avoid a importing any namespace in your global namespace.
here is how i think you should write your program.
#include <iostream>
double sum_serie(unsigned int n)
{
double prod = 1.0, sum = 0.0;
for (double c=1; c<=n ; c++)
{
prod *= ( ( 2*c ) - 1 ) / ( 2*c ); // remark the parenthesis
sum += prod;
}
return sum;
}
int main()
{
unsigned int n = 0;
int temp = 0;
std::cout << " enter the number of terms n: ";
std::cin >> temp;
if (temp > 0)
n = temp; // this is how you catch overflow
else
{
std::cout << " n < 0, no result calculated " << std::endl;
return 0;
}
std::cout << " the result is sum = " << sum_serie(n) << std::endl;
return 0;
}
I know that the question was about the implicit conversion and casting in C++, but even the way of writing a code can show you what bugs you have in it, so try to learn a proper way of expressing your ideas into code, debugging comes natural afterward.
Good Luck
I am trying to solve a question in which i need to find out the number of possible ways to make a team of two members.(note: a team can have at most two person)
After making this code, It works properly but in some test cases it shows floating point error ad i can't find out what it is exactly.
Input: 1st line : Number of test cases
2nd line: number of total person
Thank you
#include<iostream>
using namespace std;
long C(long n, long r)
{
long f[n + 1];
f[0] = 1;
for (long i = 1; i <= n; i++)
{
f[i] = i * f[i - 1];
}
return f[n] / f[r] / f[n - r];
}
int main()
{
long n, r, m,t;
cin>>t;
while(t--)
{
cin>>n;
r=1;
cout<<C(n, min(r, n - r))+1<<endl;
}
return 0;
}
You aren't getting a floating point exception. You are getting a divide by zero exception. Because your code is attempting to divide by the number 0 (which can't be done on a computer).
When you invoke C(100, 1) the main loop that initializes the f array inside C increases exponentially. Eventually, two values are multiplied such that i * f[i-1] is zero due to overflow. That leads to all the subsequent f[i] values being initialized to zero. And then the division that follows the loop is a division by zero.
Although purists on these forums will say this is undefined, here's what's really happening on most 2's complement architectures. Or at least on my computer....
At i==21:
f[20] is already equal to 2432902008176640000
21 * 2432902008176640000 overflows for 64-bit signed, and will typically become -4249290049419214848 So at this point, your program is bugged and is now in undefined behavior.
At i==66
f[65] is equal to 0x8000000000000000. So 66 * f[65] gets calculated as zero for reasons that make sense to me, but should be understood as undefined behavior.
With f[66] assigned to 0, all subsequent assignments of f[i] become zero as well. After the main loop inside C is over, the f[n-r] is zero. Hence, divide by zero error.
Update
I went back and reverse engineered your problem. It seems like your C function is just trying to compute this expression:
N!
-------------
R! * (N-R)!
Which is the "number of unique sorted combinations"
In which case instead of computing the large factorial of N!, we can reduce that expression to this:
n
[ ∏ i ]
n-r
--------------------
R!
This won't eliminate overflow, but will allow your C function to be able to take on larger values of N and R to compute the number of combinations without error.
But we can also take advantage of simple reduction before trying to do a big long factorial expression
For example, let's say we were trying to compute C(15,5). Mathematically that is:
15!
--------
10! 5!
Or as we expressed above:
1*2*3*4*5*6*7*8*9*10*11*12*13*14*15
-----------------------------------
1*2*3*4*5*6*7*8*9*10 * 1*2*3*4*5
The first 10 factors of the numerator and denominator cancel each other out:
11*12*13*14*15
-----------------------------------
1*2*3*4*5
But intuitively, you can see that "12" in the numerator is already evenly divisible by denominators 2 and 3. And that 15 in the numerator is evenly divisible by 5 in the denominator. So simple reduction can be applied:
11*2*13*14*3
-----------------------------------
1 * 4
There's even more room for greatest common divisor reduction, but this is a great start.
Let's start with a helper function that computes the product of all the values in a list.
long long multiply_vector(std::vector<int>& values)
{
long long result = 1;
for (long i : values)
{
result = result * i;
if (result < 0)
{
std::cout << "ERROR - multiply_range hit overflow" << std::endl;
return 0;
}
}
return result;
}
Not let's implement C as using the above function after doing the reduction operation
long long C(int n, int r)
{
if ((r >= n) || (n < 0) || (r < 0))
{
std::cout << "invalid parameters passed to C" << std::endl;
return 0;
}
// compute
// n!
// -------------
// r! * (n-r)!
//
// assume (r < n)
// Which maps to
// n
// [∏ i]
// n - r
// --------------------
// R!
int end = n;
int start = n - r + 1;
std::vector<int> numerators;
std::vector<int> denominators;
long long numerator = 1;
long long denominator = 1;
for (int i = start; i <= end; i++)
{
numerators.push_back(i);
}
for (int i = 2; i <= r; i++)
{
denominators.push_back(i);
}
size_t n_length = numerators.size();
size_t d_length = denominators.size();
for (size_t n = 0; n < n_length; n++)
{
int nval = numerators[n];
for (size_t d = 0; d < d_length; d++)
{
int dval = denominators[d];
if ((nval % dval) == 0)
{
denominators[d] = 1;
numerators[n] = nval / dval;
}
}
}
numerator = multiply_vector(numerators);
denominator = multiply_vector(denominators);
if ((numerator == 0) || (denominator == 0))
{
std::cout << "Giving up. Can't resolve overflow" << std::endl;
return 0;
}
long long result = numerator / denominator;
return result;
}
You are not using floating-point. And you seem to be using variable sized arrays, which is a C feature and possibly a C++ extension but not standard.
Anyway, you will get overflow and therefore undefined behaviour even for rather small values of n.
In practice the overflow will lead to array elements becoming zero for not much larger values of n.
Your code will then divide by zero and crash.
They also might have a test case like (1000000000, 999999999) which is trivial to solve, but not for your code which I bet will crash.
You don't specify what you mean by "floating point error" - I reckon you are referring to the fact that you are doing an integer division rather than a floating point one so that you will always get integers rather than floats.
int a, b;
a = 7;
b = 2;
std::cout << a / b << std::endl;
this will result in 3, not 3.5! If you want floating point result you should use floats instead like this:
float a, b;
a = 7;
b = 2;
std::cout << a / b << std::end;
So the solution to your problem would simply be to use float instead of long long int.
Note also that you are using variable sized arrays which won't work in C++ - why not use std::vector instead??
Array syntax as:
type name[size]
Note: size must a constant not a variable
Example #1:
int name[10];
Example #2:
const int asize = 10;
int name[asize];
I have a 100 digit number stored as string. I want to divide this number with an integer less than 10. How do I efficiently divide a big integer stored as a string with an integer?
You can check the big integer library.
You can use this library in a C++ program to do arithmetic on integers of size limited only by your computer's memory. The library provides BigUnsigned and BigInteger classes that represent nonnegative integers and signed integers, respectively. Most of the C++ arithmetic operators are overloaded for these classes, so big-integer calculations are as easy as:
#include "BigIntegerLibrary.hh"
BigInteger a = 65536;
cout << (a * a * a * a * a * a * a * a);
(prints 340282366920938463463374607431768211456)
Also check GMP
#WasimThabraze - what is your understanding of the longhand division method? Since the divisor is less than 1/2 the size of an integer you can use something like this for each divide:
char array[10] = {9,8,7,6,5,4,3,2,1,0};
void divide(int dvsr)
{
int rem = 0;
int dvnd;
int quot;
int i;
for(i = 0; i < (sizeof(array)/sizeof(array[0])) ; i++){
dvnd = (rem * 10) + array[i];
rem = dvnd % dvsr;
quot = dvnd / dvsr;
array[i] = quot;
}
}
int main(void)
{
divide(8);
return (0);
}
I hope this helps you because not all online judges allow BigIntegerLibrary.I have assumed for some arbitrary input.
string input="123456789";
int n=input.size();
string final(n,'0');
string::const_iterator p=input.begin(),q=input.end();
string::iterator f=final.begin();
void divide(int divisor)
{
int reminder = 0,dividend,quotient;
/*repeatedly divide each element*/
for(; p!=q ; p++,f++){
dividend = (reminder * 10) + (*p-'0');
reminder = dividend % divisor;
quotient = dividend / divisor;
*f = quotient + '0';
}
/*remove any leading zeroes from the result*/
n = final.find_first_not_of("0");
if (n != string::npos)
{
final = final.substr(n);
}
std::cout << final ;
}
int main(){
int x;
std::cin >> x;
divide(x);
return 0;
}
I wrote the following code to sum the series (-1)^i*(i/(i+1)). But when I run it I get -1 for any value of n.
Can some one please point out what I am doing wrong? Thank you in advance!
#include <iostream>
using namespace std;
int main()
{
int sum = 0;
int i = 1.0;
int n = 5.0;
for(i=1;i<=n;i++)
sum = (-1)^i*(i/(i+1));
cout << "Sum" <<" = "<< sum << endl;
return 0;
}
Problem #1: The C++ ^ operator isn't the math power operator. It's a bitwise XOR.
You should use pow() instead.
Problem #2:
You are storing floating-point types into an integer type. So the following will result in integer division (truncated division):
i/(i+1)
Problem #3:
You are not actually summing anything up:
sum = ...
should be:
sum += ...
A corrected version of the code is as follows:
double sum = 0;
int i = 1;
int n = 5;
for(i = 1; i <= n; i++)
sum += pow(-1.,(double)i) * ((double)i / (i + 1));
Although you really don't need to use pow in this case. A simple test for odd/even will do.
double sum = 0;
int i = 1;
int n = 5;
for(i = 1; i <= n; i++){
double val = (double)i / (i + 1);
if (i % 2 != 0){
val *= -1.;
}
sum += val;
}
You need too put sum += pow(-1,i)*(i/(i+1));
Otherwise you lose previous result each time.
Use pow function for pow operation.
edit : as said in other post, use double or float instead of int to avoid truncated division.
How about this
((i % 2) == 0 ? 1 : -1)
instead of
std::pow(-1, i)
?
Full answer:
double sum = 0;
int i = 1.0;
int n = 5.0;
for (i = 1; i <= n; ++i) {
signed char sign = ((i % 2) == 0 ? 1 : -1);
sum += sign * (i / (i+1));
}
Few problems:
^ is teh bitwise exclusive or in c++ not "raised to power". Use pow() method.
Remove the dangling opening bracket from the last line
Use ints not floats when assigning to ints.
You seem to have a few things wrong with your code:
using namespace std;
This is not directly related to your problem at hand, but don't ever say using namespace std; It introduces subtle bugs.
int i = 1.0;
int n = 5.0;
You are initializaing integral variables with floating-point constants. Try
int i = 1;
int n = 5;
sum = (-1)^i*(i/(i+1));
You have two problems with this expression. First, the quantity (i/(i+1)) is always zero. Remember dividing two ints rounds the result. Second, ^ doesn't do what you think it does. It is the exclusive-or operator, not the exponentiation operator. Third, ^ binds less tightly than *, so your expression is:
-1 xor (i * (i/(i+1)))
-1 xor (i * 0)
-1 xor 0
-1
^ does not do what you think it does. Also there are some other mistakes in your code.
What it should be:
#include <iostream>
#include <cmath>
int main( )
{
long sum = 0;
int i = 1;
int n = 5;
for( i = 1; i <= n; i++ )
sum += std::pow( -1.f, i ) * ( i / ( i + 1 ) );
std::cout << "Sum = " << sum << std::endl;
return 0;
}
To take a power of a value, use std::pow (see here). Also you can not assign int to a decimal value. For that you need to use float or double.
The aforementioned ^ is a bitwise-XOR, not a mark for an exponent.
Also be careful of Integer Arithmetic as you may get unexpected results. You most likely want to change your variables to either float or double.
There are a few issues with the code:
int sum = 0;
The intermediate results are not integers, this should be a double
int i = 1.0;
Since you will use this in a division, it should be a double, 1/2 is 0 if calculated in integers.
int n = 5.0;
This is an int, not a floating point value, no .0 is needed.
for(i=1;i<=n;i++)
You've already initialized i to 1, why do it again?
sum = (-1)^i*(i/(i+1));
Every iteration you lose the previous value, you should use sum+= 'new values'
Also, you don't need pow to calculate (-1)^i, all this does is switch between +1 and -1 depending on the odd/even status of i. You can do this easier with an if statement or with 2 for's, one for odd i one for even ones... Many choices really.