CF1073E.Segment Sum

You are given two integers $l$ and $r$ ($l \le r$). Your task is to calculate the sum of numbers from $l$ to $r$ (including $l$ and $r$) such that each number contains at most $k$ different digits, and print this sum modulo $998244353$.

For example, if $k = 1$ then you have to calculate all numbers from $l$ to $r$ such that each number is formed using only one digit. For $l = 10, r = 50$ the answer is $11 + 22 + 33 + 44 = 110$.

给你两个整数 $l$ 和 $r$（满足 $l \le r$）。你的任务是计算区间 $[l, r]$ 内所有满足“至多包含 $k$ 个不同数字”的整数之和，并将该和对 $998244353$ 取模后输出。

例如，若 $k = 1$，则你需要计算区间 $[l, r]$ 内所有仅由一个数字构成的整数之和。当 $l = 10, r = 50$ 时，答案为 $11 + 22 + 33 + 44 = 110$。

The only line of the input contains three integers $l$, $r$ and $k$ ($1 \le l \le r \lt 10^{18}, 1 \le k \le 10$) — the borders of the segment and the maximum number of different digits.

输入仅包含一行，三个整数 $l$、$r$ 和 $k$（$1 \le l \le r \lt 10^{18}$，$1 \le k \le 10$）—— 分别表示区间的边界以及数字中最多允许出现的不同数字的个数。

Print one integer — the sum of numbers from $l$ to $r$ such that each number contains at most $k$ different digits, modulo $998244353$.

输出一个整数——即区间 $[l, r]$ 内所有满足“其十进制表示中至多包含 $k$ 个不同数字”的整数之和，对 $998244353$ 取模。

For the first example the answer is just the sum of numbers from $l$ to $r$ which equals to $\frac{50 \cdot 51}{2} - \frac{9 \cdot 10}{2} = 1230$. This example also explained in the problem statement but for $k = 1$.

For the second example the answer is just the sum of numbers from $l$ to $r$ which equals to $\frac{2345 \cdot 2346}{2} = 2750685$.

For the third example the answer is $101 + 110 + 111 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 119 + 121 + 122 + 131 + 133 + 141 + 144 + 151 = 2189$.

第一个样例的答案即为从 $l$ 到 $r$ 的所有整数之和，其值为 $\frac{50 \cdot 51}{2} - \frac{9 \cdot 10}{2} = 1230$。该样例在题目描述中亦有说明，但对应的是 $k = 1$ 的情形。

第二个样例的答案即为从 $l$ 到 $r$ 的所有整数之和，其值为 $\frac{2345 \cdot 2346}{2} = 2750685$。

第三个样例的答案为 $101 + 110 + 111 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 119 + 121 + 122 + 131 + 133 + 141 + 144 + 151 = 2189$。