如何计算正方形序列中数字的个数

Question

There is a sequence of squares: 149162536 (1 4 9 16 25 36...) How do I get a digit that has a n-number? Example:n = 5, answer 6. n = 2, answer 4. n = 9, answer 6.

Answer 1

The solution seems rather easy with a brute force approach:

1. start with i = 0 .
2. compute i*i and convert to a string with snprintf . Let len be the length of this string.
3. if n >= len increment i , subtract len from n and continue at step 2.
4. otherwise, return the character of the string at offset n .

Here is some code:

int find_char_in_square_sequence(unsigned long long n) {
    for (unsigned long long i = 0;; i++) {
        char buf[32];
        unsigned int len = snprintf(buf, sizeof buf, "%llu", i * i);
        if (n < len)
            return buf[n] - '0';
        n -= len;
    }
}

For large index values, a more efficient approach would handle ranges of values with squares of the same size in a single step, reducing the time complexity from O(N log N) to just above O(log N) .