C++ gettinng Segmenation fault

[ankur12106037]

I was trying the question http://www.spoj.com/problems/JULKA/.
I used vector to solve it i m getting right answer on codeblocks as well as on ideone.But getting segmentation fault on spoj.
Can anyone plz tell me the possible reason for this.

#include<iostream>
#include<stdio.h>
#include<vector>
using namespace std;
vector<int> a,b,sum,minus1;
vector<int>::reverse_iterator i,j;
inline long int inputA()
{
    long int n=0;
    char c=getchar();
	while(c<'0' || c>'9')
	c=getchar();
	while(c<='9' && c>='0')
	{
      a.push_back(c-'0');
      n=(n<<1)+(n<<3)+c-'0';
      c=getchar();
	}
	return n;
}
inline long int  inputB()
{
    long int n=0;
    char c=getchar();
	while(c<'0' || c>'9')
	c=getchar();
	while(c<='9' && c>='0')
	{
      b.push_back(c-'0');
      n=(n<<1)+(n<<3)+c-'0';
      c=getchar();
	}
	return n;
}

inline void Add();

inline void sub();




int main()
{
int t=10;long int x,y;
while(t--)
{
    x=inputA();
    y=inputB();
    if(x<100)
    {
        cout<<(x+y)/2<<"\n"<<(x-y)/2;
    }
    else{
    Add();
    cout<<"\n";
    sub();
    }
    a.clear();
    b.clear();
}
return 0;
}

inline void Add()
{
    j=b.rbegin();
    i=a.rbegin();
    int c =0;
    while(j!=b.rend())
    {
        if(*i+*j+c>=10)
        {
            sum.push_back(*i+*j+c-10);
            c=1;
        }
        else
        {
            sum.push_back(*i+*j+c);
            c=0;
        }
        i++;j++;
    }
    while(c!=0||i!=a.rend())
    {
        if(*i+c>=10)
        {
            sum.push_back(*i+c-10);
            c=1;
        }
        else
        {
            sum.push_back(*i+c);
            c=0;
        }
        i++;
        if(i==a.rend()){break;}
    }
    if(c)
    {
        sum.push_back(1);
    }
     //division by 2
    int r=0,temp1;//remainder
    for(i=sum.rbegin();i!=sum.rend();i++)
    {
        if(r==0)
        {    r=(*i)%2;
             *i=*i/2;
        }
        else
        {
            temp1=(r*10+*i)%2;
            *i=(r*10+*i)/2;
            r=temp1;
        }
    }
    //for ignoring leading zeroes
    i=sum.rbegin();
    while(*i==0)
    {
        i++;
    }
    while(i!=sum.rend())
    {
        cout<<*i;
        i++;
    }
    sum.clear();
}

inline void sub()
{
    j=b.rbegin();
    i=a.rbegin();
    int c =0;
    while(j!=b.rend())
    {
        if(c==0)
        {
            if(*i-*j>=0)
            {
                minus1.push_back(*i-*j);
            }
            else
            {
                minus1.push_back(10+*i-*j);
                c=1;
            }
        }
        else
        {
            if(*i-*j-1>=0)
            {
                minus1.push_back(*i-*j-1);
                c=0;
            }
            else
            {
                minus1.push_back(9+*i-*j);
            }
        }
        i++;j++;
    }
    while(c!=0||i!=a.rend())
    {
        if(*i-c>=0)
        {
            minus1.push_back(*i-c);
            c=0;
        }
        else
        {
            minus1.push_back(9+*i);
            c=1;
        }
        i++;
    }
    //division by 2
    int r=0,temp1;//remainder
    for(i=minus1.rbegin();i!=minus1.rend();i++)
    {
        if(r==0)
        {    r=(*i)%2;
             *i=*i/2;
        }
        else
        {
            temp1=(r*10+*i)%2;
            *i=(r*10+*i)/2;
            r=temp1;
        }
    }
    //for ignoring leading zeroes
    i=minus1.rbegin();
    while(*i==0&&i!=minus1.rend())
    {
        i++;
    }
    if(i==minus1.rend()){cout<<"0";}

    else{//output
    while(i!=minus1.rend())
    {
        cout<<*i;
        i++;
    }
    }
    minus1.clear();
}

[kbw]

What are your inputs?

[dhayden]

inputA and inputB are identical except for the vector that they populate. Why not create one function that takes the vector as a parameter:
long input(vector<int> &vec);

n=(n<<1)+(n<<3)+c-'0';
It would be much clearer to say n = 10*n+c-'0';. Let the compiler decide if the optimization is worth it.

inputA() and inputB() are attempting to return the number, but the number can be as high as 10^100, which will overflow the value of a long int. Since you only care if the result is large, consider using a double, or just always using your vector to determine the answer.

Line 50: what if A<100 but B is 10^50?
Line 84 handles extra digits in A. What if the extra digits are in B, not A?
Line 165: Same problem as line 84.

[ne555]

@dhayden: B<=A

The overflow in input{A,B}() is an issue


@OP: /*165*/while(c!=0||i!=a.rend())
if the carry were in the last digit then you would try to dereference an invalid iterator