Enter an integer n to calculate the number of times 1 appears in the decimal representation of the N integers from 1 to n.

Question:

Enter an integer n to calculate the number of times 1 appears in the decimal representation of the N integers from 1 to n.

For example, input 12, integers from 1 to 12 contain numbers 1, 10, 11, and 12, and 1 appear 5 times in total.

If n = 23106 is required, the calculation process is as follows:

Tens of thousands = 1:10000 --19999 total 10000

Thousands = 1:01000--21999 for a total of 3*1000 = 3000, of which 3 indicates that 1000 bits can be filled in 1000 and 2. 3000 indicates that the last three bits have to possibilities, so there are possibilities in total.

Hundred bits = 1:The possible situation here is not: 00100--23199, because the maximum is only 23106, we can divide the sum into two parts:

00100--22199: a total of 23*100 types, of which 23 indicates thousands of digits can take 00-100 23 kinds of possibilities, 99,100 indicates that ten digits can take 00-possible.

23100--23106: a total of 7 types, that is, the thousands of digits are fixed. In only 23 cases, the ten digits can be 00-56, a total of 57 possibilities.

Therefore, the number of digits with a hundred digits being 1 is 2300 + 7 = 2307;

10 digits = 1:00010--23019 A total of 231*10 = 2310 types, of which 231 indicates that thousands of hundred digits can take 231,232-kinds of possibilities, while 10 indicates that a single digit can go 0-9 ten possibilities.

Because it takes 23110> 23106, so thousands of hundred digits = 231, and the number of digits 10 = 1 is 0

Single digit = 1:00001-- 23101A total of 2311*1

So the total number: 10000 + 3000 + 2307 + 2310 + 2311 = 19928 possibilities.

If a five-digit number is ABCDE (AB is definitely not equal to 0) and we want to calculate the number of numbers C = 1, we can discuss it in the following situations:

C> 1:00100 -- AB199, a total of (AB + 1) x 100 types, of which (AB + 1) indicates that thousands of digits can be 00-AB, a total of AB + 1 type. Because c> 1, so ab199 <ABCDE, so these numbers are within the range of 1 -- ABCDE.

C = 0:00100 -- A (b-1)199, a total of AB * 100, and A (b-1) between 200-ab099, There is no number of hundred = 1.

C = 1:We can divide the number C = 1 into the sum of the following two cases:

00100 -- A (b-1)199: A total of AB * 100 possibilities;

A (b-1)200 -- AB 0 99: the number of digits with a hundred digits = 1 is 0;

AB 1 00 -- AB 1 CD: total Cd + 1 possibility;

Therefore, the number of numbers C = 1 is equal to the number of numbers C = 0, plus ABCDE % 100 + 1.

So we only need to discuss each digit as shown above, get the number of this digit = 1, and then accumulate it to get the answer to the question lock.

#include <iostream>using namespace std; int num_of_1(int n){    /*it should cover all powers of 10 within int range*/    int pow1 = 1;    int pow2 = 10*pow1;    int  count = 0;    while(n >= pow1){//        int pow2 = 10*pow1;        switch( (n % pow2) / pow1){                case 0:                    count += (n / pow2) * pow1;                    break;                case 1:                    count += (n / pow2) * pow1;                    count += n % pow1 + 1;                    break;                default:                    count += (n / pow2 + 1) * pow1;        }        pow1 *= 10;    }    return count;} int main(){    cout << num_of_1(12) << endl;    getchar();    return 0;}