动态规划解决数字排列整除问题

问题描述

题目给定一个数字字符串 $s$ 和一个正整数 $d$，要求计算 $s$ 的所有排列中能被 $d$ 整除的排列数量。数字可能包含前导零，且如果数字相同但位置不同视为不同排列。

动态规划解法

使用状态压缩动态规划（DP）来解决。定义 $dp[mask][r]$ 表示当前已选数字的掩码为 $mask$，且当前数模 $d$ 的余数为 $r$ 时的方案数。

初始化时，$dp[0][0] = 1$，表示未选任何数字时余数为 0 的方案数为 1。

状态转移时，枚举每一位未选的数字 $s[j]$，更新新的余数： $$ new_r = (r \times 10 + (s[j] - '0')) \mod d $$

状态转移方程为： $$ dp[mask | (1 << j)][new_r] += dp[mask][r] $$

去重处理

如果数字中有重复字符，直接计算会导致重复计数。因此需要对结果除以重复字符的阶乘。具体来说，对于每个字符 $c$ 出现的次数 $cnt[c]$，最终答案为： $$ \frac{dp[full_mask][0]}{\prod_{c} cnt[c]!} $$

代码实现

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

int main() {
    int T;
    cin >> T;
    while (T--) {
        string s;
        int d;
        cin >> s >> d;
        int n = s.size();
        
        vector<vector<int>> dp(1 << n, vector<int>(d, 0));
        dp[0][0] = 1;
        
        for (int mask = 0; mask < (1 << n); ++mask) {
            for (int r = 0; r < d; ++r) {
                if (dp[mask][r] == 0) continue;
                for (int j = 0; j < n; ++j) {
                    if (!(mask & (1 << j))) {
                        int new_r = (r * 10 + (s[j] - '0')) % d;
                        dp[mask | (1 << j)][new_r] += dp[mask][r];
                    }
                }
            }
        }
        
        vector<int> cnt(10, 0);
        for (char c : s) cnt[c - '0']++;
        int fact = 1;
        for (int x : cnt) {
            for (int i = 1; i <= x; ++i) fact *= i;
        }
        
        cout << dp[(1 << n) - 1][0] / fact << endl;
    }
    return 0;
}

复杂度分析

时间复杂度为 $O(2^n \times n \times d)$，其中 $n$ 是字符串长度，$d$ 是给定的模数。空间复杂度为 $O(2^n \times d)$。适用于 $n \leq 10$ 的规模。

注意事项

• 数字可能包含前导零，但题目允许，因此无需特殊处理。
• 去重时需计算每个数字出现次数的阶乘，避免重复计数。
• 状态压缩 DP 通过掩码表示已选数字的位置，确保每个位置只选一次。

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