Project Euler Problem 32 Pandigital products

Pandigital products

Problem 32

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 ton exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

C++:

#include <iostream>
#include <algorithm>
#include <set>

using namespace std;

int main()
{
    int digits[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    long left1, left2, mid1_4, mid2_4, right4;
    set<long> s;

    for(;;) {
        left1 = digits[0];
        left2 = digits[1] + digits[0] * 10;
        mid1_4 = digits[1] * 1000 + digits[2] * 100 + digits[3] * 10 + digits[4];
        mid2_4 = digits[2] * 100 + digits[3] * 10 + digits[4];
        right4 = digits[5] * 1000 + digits[6] * 100 + digits[7] * 10 + digits[8];

        if((left1 * mid1_4 == right4) || (left2 * mid2_4 == right4))
            s.insert(right4);

        if(!next_permutation(digits, digits+9))
            break;
    }

    long total = 0;
    for(set<long>::iterator iter=s.begin(); iter!=s.end(); iter++)
        total += *iter;

    cout << total << endl;

    return 0;
}

C++:

#include <iostream>
#include <cstring>
#include <set>

using namespace std;

const long N1 = 100;
const long N2 = 2000;

bool ispandigital(long a, long b, long product)
{
    int digits[10], d;

    memset(digits, 0, sizeof(digits));

    while(a) {
        d = a % 10;
        if(digits[d])
            return false;
        digits[d] = 1;
        a /= 10;
    }
    if(digits[0])
        return false;

    while(b) {
        d = b % 10;
        if(digits[d])
            return false;
        digits[d] = 1;
        b /= 10;
    }
    if(digits[0])
        return false;

    while(product) {
        d = product % 10;
        if(digits[d])
            return false;
        digits[d] = 1;
        product /= 10;
    }
    if(digits[0])
        return false;

    for(int i=1; i<10; i++)
        if(digits[i] == 0)
            return false;

    return true;
}

int main()
{
    long long total = 0;
    long product;
    set<long> s;

    for(long i=1; i<=N1; i++)
        for(long j=i+1; j<=N2; j++) {
            product = i * j;
            if(product > 1000 && ispandigital(i, j, product))
                s.insert(product);
        }

    total = 0;
    for(set<long>::iterator iter=s.begin(); iter!=s.end(); iter++)
        total += *iter;

    cout << total << endl;

    return 0;
}