1680. Concatenation of Consecutive Binary Numbers (M)

Concatenation of Consecutive Binary Numbers (M)

题目

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 109 + 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

---

题意

将整数1-n的二进制拼成一个长二进制，求这个长二进制代表的十进制数。

思路

直接拼成字符串再计算勉强通过，也可以找到规律：(F(N)=F(N-1)<<len((N)_2)+N)。

---

代码实现

Java

class Solution {
    public int concatenatedBinary(int n) {
        long ans = 0;
        for (int i = 1; i <= n; i++) {
            int len = Integer.toBinaryString(i).length();
            ans = ((ans << len) + i) % 1000000007;
        }
        return (int)ans;
    }
}