算法

(暴搜) $O(10!)$

  send
+ more
-------
 money

$send = 1000s + 100e + 10n + d$
$more = 1000m + 100o + 10r + e$

注意到 $A + B = C$ 可以转变为 $A + B - C = 0$

所以可以在 $money$ 前加负号

$-money = -10000m - 1000o - 100n - 10e - y$

于是，

$send + more - money = 0 \Leftrightarrow 1000s + 91e - 90n + d - 9000m -900o + 10r - y$

直接枚举 $\{0, 1, 2, \dots, 9\}$ 的所有排列一个个尝试即可。

C++ 代码

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)

using std::cin;
using std::cout;
using std::map;
using std::set;
using std::vector;
using std::string;
using ll = long long;

int main() {
    vector<string> s(3);
    rep(i, 3) cin >> s[i];

    map<char, ll> mp;
    set<char> heads;
    rep(i, 3) {
        reverse(s[i].begin(), s[i].end());
        ll co = 1;
        if (i == 2) co = -1;
        for (char c : s[i]) {
            mp[c] += co;
            co *= 10;
        }
        reverse(s[i].begin(), s[i].end());
        heads.insert(s[i][0]);
    }

    if (mp.size() > 10) {
        cout << "UNSOLVABLE\n";
        return 0;
    }

    vector<int> p(10);
    iota(p.begin(), p.end(), 0);
    do{
        int i = 0;
        ll tot = 0;
        for (auto x : mp) {
            char c = x.first;
            ll co = x.second;
            if (p[i] == 0 and heads.count(c)) tot = 1e18;
            tot += co * p[i];
            ++i;
        }
        if (tot == 0) {
            i = 0;
            for (auto& x : mp) {
                x.second = p[i];
                ++i;
            }
            rep(i, 3) {
                ll x = 0;
                for (char c : s[i]) {
                    x = x * 10 + mp[c];
                }
                cout << x << '\n';
            }
            return 0;
        }
    } while (next_permutation(p.begin(), p.end()));

    cout << "UNSOLVABLE\n";

    return 0;
}