分类讨论

状态表示：$f[mid][target]$ 当前位数mid的targe出现的次数
1. 将数字放入数组 $nums： [个位…最高位]$，从高位开始枚举
2. 将数字拆分为三段：$high : mid : low$
3. 每段的区间：$high: [m, i + 1]， mid: i ，low: [i - 1, 0]$
4. 记录high所在位数：$power = 10^i$

分析target在mid位出现的次数

1. $mid == target: target$ 出现的次数为数字的值为 $high * power + low + 1$ (0 - [height 0 000] - [high target low] )
2. $mid > target : high * power + power$ (0 - [ height 0 000] - [high target 000])
3. $mid < target : mid位置不会出 现target，高位会出现：high * power$ (0 - [high 0 000])
4. 特殊情况 $target == 0$ 只统计到（ 0 - [height-1 0 000]）

C++ 代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;

int GetNum(vector<int>& nums, int l, int r){
    int res = 0;
    for(int i = l; i >= r; i--){
        res = res * 10 + nums[i];
    }
    return res;
}

int CountTimesOfTarget(int n, int target){
    if(n == 0) return 0;
    vector<int> nums;
    //先将数字存入数组
    while(n){
        nums.push_back(n % 10);
        n /= 10;
    }
    int res = 0;
    int m = nums.size() - 1;

    for(int i = m; i >= 0; i--){
        int mid = nums[i];
        int power = pow(10, i);
        int high = GetNum(nums, m, i + 1);
        int low = GetNum(nums, i - 1, 0);

        res += high * power;
        if(target == 0) res -= power;
        if(mid == target) res += low + 1;
        else if(mid > target) res += power;
    }
    return res;

}

int main(){
    int a, b;
    while(cin >> a >> b, a || b){
        if(a > b) swap(a, b);
        for(int i = 0; i < 10; i++){
            //计算i出现的次数
            if(i) cout << " ";
            cout << CountTimesOfTarget(b, i) - CountTimesOfTarget(a - 1, i);
        }
        cout << endl;
    }
}