We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. All Contests
  2. ProjectEuler+
  3. Project Euler #8: Largest product in a series
  4. Discussions

Project Euler #8: Largest product in a series

Sort by

recency

|

218 Discussions

|

  • skbenz
     + 0 comments

    Sliding window problem. You can bail on computing sub-products whenever you encounter a zero (in my case, that's when tmp becomes zero).

    int max_prod(string num, size_t n, int k) {
    int res = 0;
    size_t l = 0;
    size_t r;
        
    while (l < (n - k)) {
        r = l;
        int tmp = 1;
        while (tmp > 0 && r < l + k) {
            tmp *= (num[r] - '0');
            if ((r - l == (k - 1)) && tmp > res) {
                res = tmp;
            }
            r++;
        }
        l++;
    }
    
    return res;
}

  • skbenz
     + 0 comments

    Sliding window problem. You can bail on computing sub-products whenever you encounter a zero (in my case, that's when tmp becomes zero). int max_prod(string num, size_t n, int k) { int res = 0; size_t l = 0; size_t r;

    while (l < (n - k)) {
    r = l;
    int tmp = 1;
    while (tmp > 0 && r < l + k) {
        tmp *= (num[r] - '0');
        if ((r - l == (k - 1)) && tmp > res) {
            res = tmp;
        }
        r++;
    }
    l++;
}

return res;

    }

  • skbenz
     + 0 comments

    Sliding window problem. You can bail on computing sub-products whenever you encounter a zero (in my case, that's when tmp becomes zero).

    int max_prod(string num, size_t n, int k) { int res = 0; size_t l = 0; size_t r;

    while (l < (n - k)) {
    r = l;
    int tmp = 1;
    while (tmp > 0 && r < l + k) {
        tmp *= (num[r] - '0');
        if ((r - l == (k - 1)) && tmp > res) {
            res = tmp;
        }
        r++;
    }
    l++;
}

return res;

    }

  • skbenz
     + 0 comments

    Sliding window problem. You can bail on computing sub-products whenever you encounter a zero (in my case, that's when tmp becomes zero).

    int max_prod(string num, size_t n, int k) { int res = 0; size_t l = 0; size_t r;

    while (l < (n - k)) {
    r = l;
    int tmp = 1;
    while (tmp > 0 && r < l + k) {
        tmp *= (num[r] - '0');
        if ((r - l == (k - 1)) && tmp > res) {
            res = tmp;
        }
        r++;
    }
    l++;
}

return res;

    }

  • skbenz
     + 0 comments

    Sliding window problem. You can bail on computing sub-products whenever you encounter a zero...

    int max_prod(string num, size_t n, int k) { int res = 0; size_t l = 0; size_t r;

    while (l < (n - k)) {
    r = l;
    int tmp = 1;
    while (tmp > 0 && r < l + k) {
        tmp *= (num[r] - '0');
        if ((r - l == (k - 1)) && tmp > res) {
            res = tmp;
        }
        r++;
    }
    l++;
}

return res;

    }