I have 2 numbers, n = 1000000000000, and j = 1. When I write
cout << n / j << endl;
The console outputs the right answer, 1000000000000.
However, when I do :
int d = n / j;
cout << d << endl;
The console outputs 3567587328.
Can someone please explain why this happens and what should I do?
The value you are using is greater than the maximum value that an integer variable can store.
If you need to perform arithmetic operation with such big numbers then you will have to use special classes that handle such numbers. Perhaps your implementation of C++ supports the long long data type?
The max int you can have is 2,147,483,647 which is way smaller that 10^12 that you have there, so you have an integer overflow. Instead of an int you could use long long.
Try using long.
long long d = n / j;
cout << d << endl;
Related
I want to print the first 100 numbers in the fibonacci sequence. My program prints until around 20 numbers than the numbers turn negative.
Can someone explain this to me please and provide a fix?
Thanks,
/*Fibonacci sequence*/
#include <iostream>
using namespace std;
int main(){
long int i, fib;
int firstNum=0, secondNum=1;
cout << firstNum << endl;
cout << secondNum << endl;
for (i=0; i < 100; i++){
fib = firstNum + secondNum;
firstNum = secondNum;
secondNum = fib;
cout << fib << endl;
}
return 0;
}
What you are seeing is an integer overflow problem. firstNum and secondNum are not long.
This should fix it
unsigned long long i, fib;
unsigned long long firstNum=0, secondNum=1;
EDIT:
This will help you avoid overflow after the 20th number, but your program will still overflow. You can use unsigned long long, and you'll make it to the 100th sequence element.
Well even unsigned long long int throws out of range for higher fabonacci num(above 92) but if still interested you could store them digit by digit in array
see this
https://docs.google.com/file/d/0BwtP9e5j1RbpSjhvSG4wbkhGcmM/edit
Either We can store the values in dynamically created structures such
as Linked list (If we want to store all the Fibonacci numbers) OR
We can use just three string arrays to hold the sum and temp values to print it for this purpose which will solve your issue.
Please see this for reference
Print nth Fibonacci number [upto 1000 digits]
If you have a group of numbers, like a=4,b=5,c=8. Is there a way to get the program print out the largest value. So, for instance, I have all my values outputted 4,5,8. How can I get the program to output the largest value? (I'm using c++). I used if statements but I feel like there is a shorter way. On Google, I keep finding INT_MAX but doesn't that display the largest number in the int type?
here's part of the code
int a = (rand()%6)+1;
int b = (rand()%6)+1;
int c = (rand()%6)+1;
cout << int a << int b << int c << endl; //I'm trying to get it to display the largest int out of this group
I deleated my if statements trying to find a better way to display the largest of the group.
How can I get the program to output the largest value?
There are two steps in the process.
Compute the maximum of the three numbers.
Display the maximum number.
If you are allowed to use functions from the standard libraries, you can use std::max to compute the maximum of thee numbers. If you are not, you will have to write one yourself. Assuming you can use std::max, your code needs to be:
int m = std::max({a, b, c});
cout << "maximum: " << m << endl;
It can also be a one-liner.
cout << "maximum: " << std::max({a, b, c}) << endl;
I am adding numbers from 1 to n in C++. I have used both the iteration method and mathematical formula. The code works fine for up to 9 digits.
But when I give input a 10 digit number, the formula and iteration methods give separate answers.
I have tried to look it up on google but couldn't find any solution for this. My code:
#include <bits/stdc++.h>
using namespace std;
int main(){
unsigned long long i, n, sum = 0, out_put;
cout << "Sum of numbers from 1 to: ";
cin >> n;
/// using mathematical formula
out_put = n*(n+1);
out_put = out_put/2;
cout << " = " << out_put << endl;
/// using iteration
for (i=1; i<=n; i++){
sum = sum+i;
}
cout << " == " << sum << endl;
return 0;
}
How do know which one is the correct one? If I assume the formula can't be wrong then why is the iteration method giving incorrect answer? I have used unsigned long long to prevent overflow but still didn't work.
What you are seeing is overflow happening on your calculations at different points. 9,999,999,999 * 10,000,000,000 is ~9.9e19 while an unsigned long long holds ~1.8e19. So the result wraps around and you get one answer.
Your for loop is also going to overflow but it is going to do so at a different point meaning the answers will diverge from one another since the modulo arithmetic is happening with a smaller number.
Your problem is that n*(n+1) can be too large to store in an unsigned long long, even though the end result (half of that) which you calculate via iteration may still fit.
Assuming your unsigned long long has 64 bits, it can hold integers up to 18446744073709551615. Anything above that will restart from 0.
Edit: As Nathan points out, you can of course have both calculations overflow. The sum would still give the correct result modulo 2^64, but the direct calculation can be off because the division does not generally yield the same result modulo 2^64 after you have wrapped around.
#include <bits/stdc++.h>
using namespace std;
int main(){
unsigned long long i, n, sum = 0, out_put;
cout << "Sum of numbers from 1 to: ";
cin >> n;
/// using mathematical formula
out_put = n*(n+1);
out_put = out_put/2;
cout << " = " << out_put << endl;
/// using iteration
for (i=1; i<=n; i++){
sum = sum+i;
}
cout << " == " << sum << endl;
return 0;
}
I'm still very new to C++ still and decided to make a fibonacci sequence. It worked (Woo!) but it doesn't work as well as I would like it to.
what I mean by that is say for example I told my program to count the first 10 terms of the sequence I will get
"0, 1, 1" and then I have to press enter for each additional number until it hits ten in which case the program returns 0 and ends.
How do I get the program to display all the numbers I want to without hitting enter for each additional one?
Here is my script:
#include <iostream>
using namespace std;
int main()
{
int FibNum;
cout << "How many numbers of the Fibonacci Sequence would you like to see? \n\n";
cin>> FibNum;
cin.ignore();
int a = 0;
int b = 1;
int c = 2;
cout << "Fibonacci Sequence up to " << FibNum << " terms.\n\n";
cout << a << "\n" << b << "\n";
for (int c = 2; c < FibNum; c++) {
int d = a + b;
cout << d;
cin.ignore();
a = b;
b = d;
}
}
Thanks in advance for any help!
P.s. Also if you notice anything terrible I'm doing please feel free to correct me, I'm very aware I'm probably doing a lot wrong, I'm just trying to learn. :]
A few things:
1) Remove int c = 2; as you're re-defining c inside the for loop.
2) Drop the line cin.ignore();: in your for loop: that will fix your "enter" problem; that line waits for some input then ignores it.
3) Put some white space in your output: e.g. cout << d << ' ' so your numbers are separated.
4) [Acknowledge vincent_zhang] Consider moving to uint64_t as your data type for a, b, and d. This is a standard type in C++11. It's a 64 bit unsigned integer type; adequate for a large number of terms.
and a small thing, bordering on personal opinion,
5) Use ++c instead of c++ as the former will never run slower as, conceptually at least, post-increment has to take a copy of the original value.
Besides the previous answers,
To better format the output, add white space by changing this
cout << d;
to
cout << d << " ";
You may want to change the type of a, b and d from int to double to prevent overflow.
(If you let FibNum=100 in your code, you should be able to observe overflow, meaning that you are going to get some incorrect numbers toward the end of the sequence.)
Move cin.ignore() out of the loop then you dont need to enter to print all the 10 numbers of Fibonacci series
#include <iostream>
#include <iomanip>
using namespace std;
int a[8], e[8];
void term (int n)
{
a[0]=1;
for (int i=0; i<8; i++)
{
if (i<7)
{
a[i+1]+=(a[i]%n)*100000;
}
/* else
{
a[i+1]+=((a[i]/640)%(n/640))*100000;
}
*/
a[i]=a[i]/(n);
}
}
void sum ()
{
}
int factorial(int x, int result = 1)
{
if (x == 1)
return result;
else return factorial(x - 1, x * result);
}
int main()
{
int n=1;
for (int i=1; i<=30; i++)
{
term(n);
cout << a[0] << " "<< a[1] << " " << a[2] << " "
<< a[3] << " " << a[4] << " " << a[5]<< " "
<< " " << a[6] << " " << a[7] << endl;
n++;
for (int j=1; j<8; j++)
a[j]=0;
}
return 0;
}
That what I have above is the code that I have thus far.
the Sum and the rest are left purposely uncompleted because that is still in the building phase.
Now, I need to make an expansion of euler' number,
This is supposed to make you use series like x[n] in order to divide a result into multiple parts and use functions to calculate the results and such.
According to it,
I need to find the specific part of the Maclaurin's Expansion and calculate it.
So the X in e=1+x+(1/2!)*x and so on is always 1
Giving us e=1+1+1/2!+1/3!+1/n! to calculate
The program should calculate it in order of the N
so If N is 1 it will calculate only the corresponding factorial division part;
meaning that one part of the variable will hold the result of the calculation which will be x=1.00000000~ and the other will hold the actual sum up until now which is e=2.000000~
For N=2
x=1/2!, e=previous e+x
for N=3
x=1/3!, e=previous e+x
The maximum number of N is 29
each time the result is calculated, it needs to hold all the numbers after the dot into separate variables like x[1] x[2] x[3] until all the 30~35 digits of precision are filled with them.
so when printing out, in the case of N=2
x[0].x[1]x[2]x[3]~
should come out as
0.50000000000000000000
where x[0] should hold the value above the dot and x[1~3] would be holding the rest in 5 digits each.
Well yeah Sorry if my explanation sucks but This is what its asking.
All the arrays must be in Int and I cannot use others
And I cant use bigint as it defeats the purpose
The other problem I have is, while doing the operations, it goes well till the 7th.
Starting from the 8th and so on It wont continue without giving me negative numbers.
for N=8
It should be 00002480158730158730158730.
Instead I get 00002 48015 -19220 -41904 30331 53015 -19220
That is obviously due to int's limit and since at that part it does
1936000000%40320
in order to get a[3]'s value which then is 35200 which is then multiplied by 100000
giving us a 3520000000/40320, though the value of a[3] exceeds the limit of integer, any way to fix this?
I cannot use doubles or Bigints for this so if anyone has a workaround for this, it would be appreciated.
You cannot use floating point or bigint, but what about other compiler intrinsic integral types like long long, unsigned long long, etc.? To make it explicit you could use <stdint.h>'s int64_t and uint64_t (or <cstdint>'s std::int64_t and std::uint64_t, though this header is not officially standard yet but is supported on many compilers).
I don't know if this is of any use, but you can find the code I wrote to calculate Euler's number here: http://41j.com/blog/2011/10/program-for-calculating-e/
32bit int limits fact to 11!
so you have to store all the above facts divided by some number
12!/10000
13!/10000
when it does not fit anymore use 10000^2 and so on
when using the division result is just shifted to next four decimals ... (as i assumed was firstly intended)
of course you do not divide 1/n!
on integers that will be zero instead divide 10000
but that limits the n! to only 9999 so if you want more add zeroes everywhere and the result are decimals
also i think there can be some overflow so you should also carry on to upper digits