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up vote 3 down vote favorite

This is my solution to the competitive coding question on SPOJ here. It gives a runtime of 0.24 but the best solutions posted have a 0.0 runtime. How could I possibly achieve better results than this?

#include<iostream>
#include<cstdio>

using namespace std;

int rev(int num)
{
    int rev_num=0;
    while(num)
    {
        rev_num= rev_num*10 + num%10;
        num/=10;
    }
    return rev_num;
}
int main()
{   
    int n, n1, n2;
    scanf("%d",&n);
    while(n--)
    {
        cin>>n1>>n2;
        cout<<rev(rev(n1)+rev(n2))<<endl;

    }
    return 0;
}
share|improve this question
    
Nice first post, welcome to Code Review! –  Mat's Mug Jun 18 at 16:59

1 Answer 1

up vote 1 down vote

A faster method would avoid the numeric divisions. Instead, it would load the number in a string (through the sprintf function), reverse the string, and finally recast the string into a number (with the help of the sscanf function).

I'm attaching a function named fastrev which in turn uses an auxilliary function reverse_string. Hopefully, this would improve runtime for large numbers.

char* reverse_string(char *s1, char* s2)
{
    char *p, *q;
    int len = strlen(s1);

    for (p = s1; *p; p++) {
        s2[len-1] = *p;
        len--;
    }
    return s2;
}

int fastrev(int num) {
    static char nbuff[32];
    static char rnbuff[32];
    int rnum;

    sprintf(nbuff, "%d", num);
    reverse_string(nbuff, rnbuff);
    sscanf(rnbuff, "%d", &rnum);
    return rnum;
}
share|improve this answer
    
's2[len] = '\0';' is needed before the for loop in the reverse_string function to reset s2 before each use so that the trailing characters from the previous reversal are not considered this time. Other than that, good alternative. But it increases the time to 0.38. I think all the memeory read/write of the char arrays is to blame. –  Antigone93 Jun 18 at 21:03
    
Actually no... Recall that s2 points to the rnbuff which is statically allocated and hence initialized to all zeroes. Memory read/write shouldn't be an issue because it is very fast... The Disk I/O is expected to be slow... How exactly are you measuring the runtime? If you want to squeeze runtime to the extreme possible extent, you might swap a[0] with a[n-1], a[1] with a[n-2] and so on... then you wont require the second buffer.... –  Debasis Jun 18 at 21:33

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