[leetcode] 1680. Concatenation of Consecutive Binary Numbers

[leetcode] 1680. Concatenation of Consecutive Binary Numbers

Description

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

Example 2:

Input: n = 3Output: 27Explanation: 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 = 12Output: 505379714Explanation: The concatenation results in "1101110010111011110001001101010111100".The decimal value of that is 118505380540.After modulo 109 + 7, the result is 505379714.

Constraints:

1 <= n <= 105

分析

题目的意思是：给定一个数n，求出把1～n表示成二进制然后拼接一起的数。这道题把它转换成10进制来做，首先要确定对于数i，需要把数向左移动的长度，这里用base，当i&(i-1)==0的时候，说明移动的长度base要更新了，更新base了之后再进行计算。 比如： `` 1 1 2 10 3 11 4 100 5 101 …

其中遍历到2，4，8...的时候需要更新向左移动的长度，即需要更新base了。res的计算为：

res=((res<

为避免数过大，需要进行取余操作## 代码

class Solution: def concatenatedBinary(self, n: int) -> int: mod=10**9+7 base=0 res=0 for i in range(1,n+1): if(i&(i-1)==0): base+=1 res=((res<

## 参考文献[连接连续二进制数字](https://leetcode-cn.com/problems/concatenation-of-consecutive-binary-numbers/solution/lian-jie-lian-xu-er-jin-zhi-shu-zi-by-ze-t40j/)

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