Count 1 between 0 and N

Descriptive narrative of the problem
Given a decimal integer n, the number of "1" appears in all integers from 1 to n.

For example: 1 "1" appeared in n=2.

n=12 when 1,2,3,4,5,6,7,8,9,10,11,12.

There were 5 "1".

Thinking of solving problems

1-digit situation:

In the solution two has been analyzed, greater than or equal to 1 times. There are 1, less than 1 there is no.

2-digit situation:

N=13, the number of single-digit occurrences of 1 is 2. 1 and 11, 10 digits 1 are 4, respectively 10,11,12,13, so f (N) = 2+4.

N=23, the number of single-digit occurrences of 1 is 3, respectively, 1,11,21, 10 digits 1 of the number of 10, respectively, 10~19,f (N) =3+10.

As a result, we find. Single-digit occurrences of 1 are not only related to single-digit numbers. and 10 digits are also concerned, assuming that the single digit is greater than or equal to 1, the number of single-digit 1 is 10-digit number plus 1, assuming that the single digit is 0. Single-digit occurrences of 1 are equal to 10-digit numbers. The number of 10-digit occurrences of 1 is not only related to the 10-digit number, but also to the single-digit: Assuming that the 10-digit number equals 1, 10 on the 1-digit number is the digit of the single digit plus 1, if the 10-digit number is greater than 1, 10 is 1 on the 10-digit number.

3-digit situation:

N=123

The number of single-digit occurrences of 1 is 13:1,11,21, .... 91,101,111,121

The number of 10-bit occurrences of 1 is 20:10~19,110~119

The number of hundreds of 1 appears to be 24:100~123

We can continue to analyze 4-digit, 5-digit numbers and derive the following general conditions:

Suppose N, we want to calculate the number of occurrences of 1 on the Hundred. will be decided by three parts: the number on the hundred, the number above the hundred. Hundreds of numbers.

Assuming that the number on the hundred is 0, then the number of occurrences of 1 on the hundred is determined only by a higher level, for example 12013, where 1 of the hundreds appear as 100~199,1100~1199,2100~2199,...,11100~11199, a total of 1200. equals the higher number multiplied by the current number of digits. That is 12 * 100.

Assuming that the number on the hundred is greater than 1, the number of occurrences of 1 on the hundred is determined only by the higher. For example, 12213, hundreds of 1 of cases are 100~199,1100~1199,2100~2199. ..., 11100~11199. 12100~12199 a total of 1300.

equals the higher number plus 1 times the current number of digits, i.e. (12 + 1) *100.

Assuming that the number on the hundred is 1, the number of occurrences of 1 on hundreds is not only affected by higher levels, but also by low levels. For example, 12113, is affected by high level 1 of the situation: 100~199,1100~1199,2100~2199. ..., 11100~11199, a total of 1200, but it is also affected by low, 1 of the situation is 12100~12113, a total of 114, equal to the low number 113+1.

Implementation code

#include <iostream>#include <cstring>using namespace STD;intCountone (intN) {intCurrent =0;intBefore =0;intafter =0;inti =1;intCount =0; while(n/i! =0) {current = n/i%Ten; Before = n/(i *Ten); after = N-(n/i) * i;if(Current >1) {count + = (before +1) * I; }Else if(Current = =0) {count + = before * I; }Else if(Current = =1) {count + = before * i + after +1; } I *=Ten; }returnCount;}intMain () {intN while(Cin>>n) {cout<<countone (n) <<endl; }return 0;}

Count 1 between 0 and N