1.和leetcode里面的Number of Digit One一样，统计0~n中，出现多少个1.
2.算法说明：
如3141592，在m（digitDivide）=100时，即要求计算百位上“1”的个数
其中a为31415，b为92，31415中出现了3142次“1”，因为每10个数里面出现1次“1”。而实际上，31415是3141500，所以把31415中1的个数再乘以m。如3141400~3141499中，前缀为31414的数出现了100次，所以需要乘以m（此时是100）。
The task is simple: given any positive integer N, you are supposed to count the total number of 1’s in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1’s in 1, 10, 11, and 12.
Input Specification:
Each input file contains one test case which gives the positive N (<=230).
Output Specification:
For each test case, print the number of 1’s in one line.
Sample Input:
12
Sample Output:
5
AC代码：
[c language=”++”]
#include <iostream>
#include <stdio.h>
#include <vector>
#include <stack>
#include <algorithm>
#include <memory.h>
#include <map>
#include <set>
#include "limits.h"
using namespace std;
int main(void)
{
long long n;
cin>>n;
long long ans = 0;
for(long long digitDivide=1;digitDivide<=n;digitDivide*=10)
{
long long a = n / digitDivide;
long long b = n%digitDivide;
ans+=(a+8)/10*digitDivide+(a%10==1)*(b+1);
}
cout<<ans<<endl;
return 0;
}
[/c]