1.number of digit one 升级版
2.用 f(high)-f(low-1)
3.digit 0 的case 要额外注意，其他就是可以用数学堆出来



Given an integer d between 0 and 9, and two positive integers low and high as lower and upper bounds, respectively. Return the number of times that d occurs as a digit in all integers between low and high, including the bounds low and high.
 

Example 1:

Input: d = 1, low = 1, high = 13
Output: 6
Explanation: 
The digit d=1 occurs 6 times in 1,10,11,12,13. Note that the digit d=1 occurs twice in the number 11.
Example 2:

Input: d = 3, low = 100, high = 250
Output: 35
Explanation: 
The digit d=3 occurs 35 times in 103,113,123,130,131,...,238,239,243.
 

Note:

0 <= d <= 9
1 <= low <= high <= 2×10^8








class Solution {
public:
    int digitsCount(int d, int low, int high) {
        return calculate(high,d)-calculate(low-1,d);
    }
    
    int calculate(int n,int d){//n is the bound => [0,n] => how many digit d occur
        if(n==0&&d==0)return 1;
        if(n<d)return 0;
        int res=0;
        string num=to_string(n);
        
        vector<int>A;
        A.push_back(1l);
        int mul=10;
        for(int i=1;i<9;i++){
            int sum=A[A.size()-1]*10+mul;
            A.push_back(sum);
            mul*=10;
        }
        
        mul=1;
        for(int i=0;i<num.size()-1;i++){
            mul*=10;
        }
        int index=num.size()-2;
        
        cout<<mul<<endl;
        
        for(int i=0;i<num.size();i++){
            int digit=num[i]-'0';
            if(i==num.size()-1){
                if(digit>=d){
                    res++;
                }
                break;
            }
            if(d==0&&i==0){
                res+=((digit)*A[index]);
                int div=10;
                for(int j=0;j<num.size()-2;j++){
                    res-=(mul/div);
                    div*=10;
                }
                mul/=10;
                index--;
                continue;
            }
            
            if(digit>d){
                res+=((digit)*A[index]+mul); 
            }
            else if(digit<d){
                res+=((digit)*A[index]);
            }
            else{
                res+=((digit)*A[index]+stoi(num.substr(i+1))+1);
            }
            mul/=10;
            index--;
        }
        return res;   
    }
    
};