AcWing 338. 计数问题 数位dp

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

using namespace std;

const int N = 10;

/*

001~abc-1, 999

abcxegf
    1. num[i] < x, 0
    2. num[i] == x, 0~efg
    3. num[i] > x, 0~999

*/

int get(vector<int> num, int l, int r) {//求前面这些位组成的数字是多少
    int res = 0;
    for (int i = l; i >= r; i -- ) res = res * 10 + num[i];
    return res;
}

int power10(int x) {//求10的i次方
    int res = 1;
    while (x -- ) res *= 10;
    return res;
}

int count(int n, int x) {//1到n当中，x出现的次数
    if (!n) return 0;
    vector<int> num;//每一位
    while (n) {
        num.push_back(n % 10);
        n /= 10;
    }
    n = num.size();
    int res = 0;
    for (int i = n - 1 - !x ; i >= 0; i -- ) {
        //判断是否是在枚举最高位
        if (i < n - 1) {
            //            i前面所有这些位数
            res += get(num, n - 1, i + 1) * power10(i);
            if (!x) res -= power10(i);
        }
        if (num[i] == x) res += get(num, i - 1, 0) + 1;
        else if (num[i] > x) res += power10(i);
    }

    return res;
}

int main() {
    int a, b;
    while (cin >> a >> b , a) {
        if (a > b) swap(a, b);

        for (int i = 0; i <= 9; i ++ )
            cout << count(b, i) - count(a - 1, i) << ' ';
        cout << endl;
    }

    return 0;
}