[LeetCode January Challange] Day 27 - Concatenation of Consecutive Binary Numbers

Given an integer n, return *the **decimal value** of the binary string formed by concatenating the binary representations of 1 to n in order, **modulo** 10^9 + 7*.

Example 1:

Input: n = 1
Output: 1
Explanation: "1" in binary corresponds to the decimal value 1. 

Example 2:

Input: n = 3
Output: 27
Explanation: In binary, 1, 2, and 3 corresponds to "1", "10", and "11".
After concatenating them, we have "11011", which corresponds to the decimal value 27.

Example 3:

Input: n = 12
Output: 505379714
Explanation: The concatenation results in "1101110010111011110001001101010111100".
The decimal value of that is 118505380540.
After modulo 109 + 7, the result is 505379714.

Constraints:

• 1 <= n <= 10^5

---

Solution

Time complexity : O(n)
Space complexity : O(1)

class Solution {
public:
    int concatenatedBinary(int n) {
        constexpr int MOD = 1e9+7;
        unsigned long ans = 0;
        for (int i=1, len=0; i<=n; ++i) {
            if ((i&(i-1))==0) ++len;
            ans = ((ans << len) % MOD + i) % MOD;
        }
        return ans;
    }
};

len 指的是 i 用 2 進位表示的長度。
ans 從 1 開始，左移 i 所需的 bit，取 mod 後再加上 i。