1680. Concatenation of Consecutive Binary Numbers

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

思路：

首先如果是真的按照思路走用字符串拼接的话肯定会超时，因为数据规模在10^5，只能 $O(n)$ 或者更小。

根据观察得出规律：

n=1, f(1)=1

n=2, f(2)=[f(1) << 2] + 2 = 6

n=3, f(3)=[f(2) << 2] + 3 = 27

n=4, f(4)=[f(3) << 3] + 4 = 220

可以看出左移的位数和n的位数是相同的。

因此手动计算当前n的位数，然后进行左移和加n，中间不要忘记取模。另外因为数值过大，最初的记录答案的类型可以为long long。

代码：

class Solution {
public:
int concatenatedBinary(int n) {
int mode =1e9+7;
long long ans=1;
for(int i=2;i<=n;i++)
{
ans=ans<<count(i);
ans%=mode;
ans+=i;
ans%=mode;
}
return ans;
}

private:
int count(int n)
{
int ans=0;
while(n>0)
{
ans++;
n=n>>1;
}
return ans;
}
};

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