虫食算

1.严格按本位加法等式推导未知字母，数学推导无漏洞；
2.数字从 n-1 倒序枚举，加速找到唯一解，解决 TLE；
3.输入加速 ios::sync_with_stdio(false); cin.tie(0); 降低常数
进位只存在 0/1，推导循环仅 k=0、1，无多余分支
代码如下：

#include <iostream>
#include <cstring>
#include <string>
using namespace std;
int n, p, a, b, c, va, vb, vc, total, want_a, want_b, want_c, out_carry, tar;
string s1, s2, s3;
int num[26];
bool used[26], ans_found;
void dfs(int pos, int in_carry){
    if (ans_found){
        return;
    }
    if (pos == n){
        if (in_carry == 0) {
            ans_found = true;
        }
        return;
    }
    p = n - 1 - pos;
    a = s1[p] - 'A';
    b = s2[p] - 'A';
    c = s3[p] - 'A';
    va = num[a];
	vb = num[b];
	vc = num[c];
    if (va != -1 && vb != -1 && vc != -1){
        total = va + vb + in_carry;
        if (total % n == vc){
            dfs(pos + 1, total / n);
        }    
        return;
    }
    if (va != -1 && vb != -1){
        total = va + vb + in_carry;
        want_c = total % n;
        out_carry = total / n;
        if (vc == -1 && !used[want_c]){
            num[c] = want_c;
            used[want_c] = true;
            dfs(pos + 1, out_carry);
            if (ans_found){
                return;
            } 
            num[c] = -1;
            used[want_c] = false;
        }
        return;
    }
    if (va != -1 && vc != -1){
        for (int k = 0; k <= 1; k++){
            want_b = vc + k * n - va - in_carry;
            if (want_b >= 0 && want_b < n && vb == -1 && !used[want_b]){
                num[b] = want_b;
                used[want_b] = true;
                dfs(pos + 1, k);
                if (ans_found) return;
                num[b] = -1;
                used[want_b] = false;
            }
        }
        return;
    }
    if (vb != -1 && vc != -1){
        for (int k = 0; k <= 1; k++){
            want_a = vc + k * n - vb - in_carry;
            if (want_a >= 0 && want_a < n && va == -1 && !used[want_a]){
                num[a] = want_a;
                used[want_a] = true;
                dfs(pos + 1, k);
                if (ans_found) {
                    return;
                }
                num[a] = -1;
                used[want_a] = false;
            }
        }
        return;
    }
    if (va == -1) {
        tar = a;
    }
    else{
        if (vb == -1){
             tar = b;
        }
        else{
            tar = c;
        }
    } 
    for (int d = n - 1; d >= 0; d--) {
        if (used[d]){
            continue;
        } 
        num[tar] = d;
        used[d] = true;
        dfs(pos, in_carry);
        if (ans_found){
            return;
        } 
        num[tar] = -1;
        used[d] = false;
    }
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> s1 >> s2 >> s3;
    memset(num, -1, sizeof(num));
    memset(used, 0, sizeof(used));
    ans_found = false;
    dfs(0, 0);
    for (int i = 0; i < n; i++){
        cout << num[i];
        if (i != n - 1) {
            cout << " ";
        }
    }
    return 0;
}