这道题花了我一个下午的时间才ac，隔壁的@zhangjianjunab 大佬一个小时就切了%%%%%%%
状态定义：f[位数][总和][模数][%完以后剩余的数]
然后求1-x的月之数
从高到低枚举每一位应该填的是，如果这一位x的数字是a[i]，那么我们就从0枚举到a[i]-1，直接统计个数。
对于当前填到a[i]，我们就不计算，用一个数组存起来，影响以后的统计

#include <cstdio>
#define LL long long
using namespace std;

LL f[11][83][83][83], ten[11];
//f[i][j][x][y] 有i位， 和为j，%x=y 的数的个数

int len, a[11];

LL solve(LL x) {
    if (!x) return 0;
    len = 0; LL t = x, s = 0;
    while (x) { 
        a[++len] = int(x % 10);
        s += a[len];
        x /= 10;
    }
    LL ans = 0;
    if (t % s == 0) ans++;
    for (int j = 1; j <= 82; j++) {
        int sum = j; LL now = 0;
        for (int i = len; i >= 1; i--) {
            for (int k = 0; k < a[i]; k++)
                if (k <= sum)
                    ans += f[i - 1][sum - k][j][(j - (now + k * ten[i - 1] % j) % j) % j];
            sum -= a[i]; if (sum < 0) break; 
            now = (now + a[i] * ten[i - 1] % j) % j;
        }
    }
    return ans;
}

int main() {
    ten[0] = 1; for (int i = 1; i <= 10; i++) ten[i] = ten[i - 1] * 10;
    for (int i = 1; i <= 82; i++) f[0][0][i][0] = 1;
    for (int i = 1; i <= 10; i++)
        for (int j = 0; j <= 82; j++)
            for (int x = 1; x <= 82; x++)
                for (int y = 0; y < x; y++) 
                    for (int num = 0; num <= 9; num++)
                        if (num <= j) 
                            f[i][j][x][y] += f[i - 1][j - num][x][(y - num * ten[i - 1] % x + x) % x];
    LL x, y;
    while (~scanf("%lld%lld", &x, &y))
        printf("%lld\n", solve(y) - solve(x - 1));
    return 0;
}