CODEWITHSUNDEEP
cproject
user3178186
user3178186

Reputation: 3

Solving Project Euler #4 with C language

I tried to solve Problem 4 of Project Euler with C language but get wrong answers all the time.

// Project Euler - Problem 5 
// 09/01/2014
#include <stdio.h>

int a,b,c,digits[14],e,y,z,biggestNum;

void isPalindrome (int x)
{
    a = -1;
    c = 0;
    b = x;
    while (b != 0)
    {
        digits[c] = (b % 10);
        b=b/10;
        c++;
    }
    while (c>=a)
    {
        if(digits[++a]!=digits[--c])
        {
            break;
        }
        if(a==c) {  biggestNum=x; }
        else if(a==c-1) {   biggestNum=x; }
    }
}

int main (void)
{
    for(y=10; y<1000; y++)
    {
        for(z=10; z<1000; z++)
        {
            isPalindrome(y*z);
        }
    }
    printf ("%d",biggestNum);
    return 0;
}

Can someone tell me what is wrong in my code? Maybe in checking if the num is Palindrome function? thanks The Problem

Upvotes: 0

Views: 894

Answers (3)

JeremyP
JeremyP

Reputation: 86691

Your code doesn't find the biggest Palindrome, it finds the last Palindrome if you run through the numbers 

10* 10, 10 * 11, 10 * 12, .... 10 * 999
...
999 * 10, 999 * 11, ....       999 * 999

If the largest palindrome happens to be (say) 750 * 750, but also 999 * 11 is a Palindrome, the 999 * 11 will overwrite the correct answer.

You need to test each time that your answer is bigger than the previous biggest answer you had.

Upvotes: 1

BLUEPIXY
BLUEPIXY

Reputation: 40155

change

    while (c>=a)
    {
        if(digits[++a]!=digits[--c])
        {
            break;
        }
        if(a==c) {  biggestNum=x; }
        else if(a==c-1) {   biggestNum=x; }
    }

to

    while (c > a){
        if(digits[++a]!=digits[--c])
            return ;
    }
    if(x>biggestNum)//bigger than old value
        biggestNum=x;

Upvotes: 0

ichramm
ichramm

Reputation: 6642

  1. You can improve the foor loop with the following:
    for(y = 999; y >= 100 ; --y)
    {
        for(z = y; z >= 100; --z)
        {
            if (isPalindrome(y*z)) {
                return y*z; // TODO: use temp var
            }
        }
    }
  1. It will be easier to convert the number to string and the check using n/2 steps (being n the number of digits)
    for (i = 0; i < n/2; i++)
    {
        if (str[i] != str(n-i)) {
            return false;
        }
    }
    return true

Upvotes: 1

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