Educational Codeforces Round 53 (Rated for Div. 2) E. Segment Sum 题解 数位dp

Segment Sum

题目描述

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 $11+22+33+44=110$.

输入格式

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

输出格式

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

提示

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 $101+110+111+112+113+114+115+116+117+118+119+121+122+131+133+141+144+151=2189$.

题面翻译

给定 $l,r,k$，求 $[l,r]$ 之间的各数位上包含的不同数字不超过 $k$ 个的所有数的和。答案对 $998244353$ 取模。

数据规模与约定

$1\le l\le r\le 10^{18}$，
$1\le k\le 10$。

样例 #1

样例输入 #1

10 50 2

样例输出 #1

1230

样例 #2

样例输入 #2

1 2345 10

样例输出 #2

2750685

样例 #3

样例输入 #3

101 154 2

样例输出 #3

2189

原题

CF——传送门
洛谷——传送门

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll mod = 998244353;
int k;                              // 条件：不超过k个不同数字
pair<int, int> dp[20][1 << 12][12]; // dp记录[pos][mask][last]状态下的满足条件的个数
int a[20];                          // a[]记录数字串

// 检查此时是否超过了k个不同数字
bool check(ll mask)
{
    int tot = 0;
    for (int i = 0; i <= 9; i++)
    {
        if ((mask >> i) & 1)
            tot++;
    }
    if (tot > k)
        return 0;
    else
        return 1;
}

// 进行答案的运算
void add(pair<int, int> &res, pair<int, int> dfs_ret, int num, int pos)
{
    res.first += dfs_ret.first;
    res.first %= mod;

    res.second += 1ll * dfs_ret.first % mod * num % mod * ((ll)pow(10, pos) % mod) % mod;
    res.second %= mod;
    res.second += dfs_ret.second;
    res.second %= mod;
}

// pos为此时的位置，mask是对0~10各个数是否存在的状态压缩，last记录上一个数，bound表示前面每一位是否都是上界，lead_0表示是否前面全是0
pair<int, int> dfs(int pos, ll mask, int last, bool bound, bool lead_0) // pair<int,int>的first记录当前方案数，second记录当前总和
{
    if (!check(mask))
        return {0, 0};
    if (pos == 0)                                             // 枚举完每一位时返回
        return {1, last};                                     // 方案数+1，总和+last
    if (!bound && !lead_0 && dp[pos][mask][last].first != -1) // 读取记忆(doge
        return dp[pos][mask][last];
    int max_num; // 可枚举的该位的数的上界
    if (bound)
        max_num = a[pos];
    else
        max_num = 9;
    pair<int, int> res = {0, 0}; // 统计方案数和总和
    for (int i = 0; i <= max_num; i++)
    {
        if (lead_0 && i == 0)
            add(res, dfs(pos - 1, mask, last, bound && (i == a[pos]), 1), 0, 0);
        else
            add(res, dfs(pos - 1, mask | (1 << i), i, bound && (i == a[pos]), 0), last, pos);
    }
    if (!bound && !lead_0) // 没在边界时，记录下该状态对应的答案，记忆化
        dp[pos][mask][last] = res;
    return res;
}

int solve(ll x)
{
    int len = 0; // 数字长度
    while (x)
    {
        a[++len] = x % 10;
        x /= 10;
    }
    return dfs(len, 0, 0, 1, 1).second;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    memset(dp, -1, sizeof(dp)); // 将dp数组初始化为-1，表示对应状态的答案目前还未计算出
    ll l, r;
    cin >> l >> r >> k;
    cout << ((solve(r) - solve(l - 1)) % mod + mod) % mod << '\n';

    return 0;
}