This afternoon, I was bored in my entrepreneurship education class. I thought of a Microsoft interview question ~ Number of times 1 appears in N

For example, if n = 11, then 1 appears four times: 1 10 11.

I have been thinking about this issue for a while and launched a fast solution (complexity lg (N )...

Idea: first determine the number of single-digit 1, and then count the number of ten-digit 1.

The procedure for setting n = 497 is as follows:

One digit:

Because the "49" in 497 is obtained by the bitwise carry, it is not difficult to come up with a single digit that has a forward of 49 digits. Each carry digit has a 1, therefore, the number of a single digit 1 is equal to 490/10 + the remaining '7' contains 1 digit, totaling 490/10 + 1 = 50

10 to 1:

It is not hard to think that the numbers above 497 are all generated by the decimal places. Therefore, the decimal places are 400/100 = four times, and each decimal place contains 10 digits of a single digit, therefore, the number of "1" in 10 of 497 is 400/100x10 + 10 = 50.

Hundred bits and 1 bits: Because the hundred bits are the highest bits and the hundred bits are greater than 1, the hundred bits contain 1 bits and 100 bits...

When N = 497 is accumulated, the total number of 1 instances is 50 + 50 + 100 = 200 .....

Based on the above idea, we can get an algorithm with the complexity of lg (n)... implementation as follows:

# Include <iostream> <br/> # include <math. h> <br/> using namespace STD; </P> <p> int get (int n) // returns 1 ~ The number of times that 1 appears in n. The code is concise <br/>{< br/> int lenth = (INT) (log10 (N) + 1, I, j, k = 1, sum = 0; <br/> for (I = 1; I <= lenth; I ++) <br/> {<br/> K * = 10; <br/> sum + = N/K * k/10; <br/> J = (N % K) * 10/K; <br/> If (j> 1) sum + = K/10; <br/> If (j = 1) sum + = (N % K) % (K/10) + 1; <br/>}< br/> return sum; <br/>}</P> <p> int main () <br/> {<br/> int N; <br/> while (CIN> N & N) <br/> cout <get (n) <Endl; <br/>}< br/>