Why is it showing floating-point exception (SIGFPE).

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The problem is to calculate Combinations(nCr). user has to input no of test cases , the number n and number r.

Input:
First line contains number of test cases T. T testcases follow. Each testcase contains 1 line of input containing 2 integers n and r separated by a space.

Output:
For each testcase, in a new line, find the nCr. Modulus your output to 109+7.

Constraints:
1 <= T <= 50
1<= n <= 103
1<= r <=800 ...

What I have tried:

C++

#include <bits/stdc++.h>
using namespace std;

int main() {
int t;
cin>>t;
while(t--)
{
    int n,r;
    cin>>n>>r;
    if(r>n)
    cout<<"0"<<endl;
    else if(r==1)
    cout<<n<<endl;
    else if(r==n)
    cout<<"1"<<endl;
    else if(n-r==1)
    cout<<n<<endl;
    else
    {
        int i,j;
        long  int res;
        long  int num=n;
        for(i=n-1;i>=(n-r+1);i--)
            num*=i;
            long int ch=(pow(10,9)+7);
        long  int den=r;
        for(i=r-1;i>=2;i--)
           den*=i;
           if((num/den)<ch)
           res=(num/den);
           else if ((num/den)==ch)
           res=0;
           else
           res=((num/den)%ch);
           cout<<res<<endl;
    }
}
	return 0;
}

Posted 14-Jan-19 7:55am

Prateek Krishna

Updated 15-Jan-19 1:36am

KarstenK

v2

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Comments

Rick York 14-Jan-19 15:24pm

One thing that can help code readability a lot is to properly indent your code. Then you can see the hierarchy and extent of loops better. In your code is difficult to see where you mean for the for loops to extend. Just like ifs and elses, the for loop extends for one statement unless it has curly braces. It appears to me that your for loops need some braces added. It is a good habit to always add them because then it will be easier when you have to add statements to the loop.

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Patrice T · Answer 1 · 2019-01-14T10:10:00

Learn to indent properly your code, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.

C++

#include <bits/stdc++.h>
using namespace std;

int main() {
    int t;
    cin>>t;
    while(t--)
    {
        int n,r;
        cin>>n>>r;
        if(r>n)
            cout<<"0"<<endl;
        else if(r==1)
            cout<<n<<endl;
        else if(r==n)
            cout<<"1"<<endl;
        else if(n-r==1)
            cout<<n<<endl;
        else
        {
            int i,j;
            long  int res;
            long  int num=n;
            for(i=n-1;i>=(n-r+1);i--)
                num*=i;
            long int ch=(pow(10,9)+7);
            long  int den=r;
            for(i=r-1;i>=2;i--)
               den*=i;
           if((num/den)<ch)
               res=(num/den);
           else if ((num/den)==ch)
               res=0;
           else
               res=((num/den)%ch);
           cout<<res<<endl;
        }
    }
        return 0;
}

Indentation style - Wikipedia[^]

Professional programmer's editors have this feature and others ones such as parenthesis matching and syntax highlighting.
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Quote:
Why is it showing floating-point exception (SIGFPE).

Because your algorithm lead to huge integers that are beyond the limit of integers.
- Some refinement in your algorithm can prevent integers from growing too fast. study the problem with examples: take a set of 10 pieces and see how the maths works if you take 1, 2, 3 ... . Compare results, how you get the answer ...
- Using bigint (infinite integers) is also a solution.

OriginalGriff · Answer 2 · 2019-01-14T08:16:00

Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.

So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.

Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:

Input   Expected output    Actual output
  1            2                 1
  2            4                 4
  3            6                 9
  4            8                16

Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:

C#

int Double(int value)
   {
   return value * value;
   }

Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on the first line of the function, and run your app. When it reaches the breakpoint, the debugger will stop, and hand control over to you. You can now run your code line-by-line (called "single stepping") and look at (or even change) variable contents as necessary (heck, you can even change the code and try again if you need to).
Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
Hopefully, that should help you locate which part of that code has a problem, and what the problem is.
This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!

KarstenK · Answer 3 · 2019-01-15T01:37:00

Solution 3

The reason of a SIGPFE is a bug in the code, often is it a division with 0 as divisor.

I recommand using the debugger and test to 0 before divising.

Posted 15-Jan-19 1:37am

KarstenK

Why is it showing floating-point exception (SIGFPE).

3 solutions

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