7-127 4010 古代密码

对每一位枚举两个距离，并且将每个距离生成两个字符串，对生成的结果以及原字符串排序比较，如果有完全一致的情况，则判定为 YES ，反之为 NO 。

#include <algorithm>
#include <cstdint>
#include <iostream>
#include <limits>
#include <numeric>
#include <string>
#include <unordered_map>
#include <utility>
#include <vector>

using i32 = std::int32_t;
using i64 = std::int64_t;
using u32 = std::uint32_t;
using u64 = std::uint64_t;
using pii = std::pair<i32, i32>;
using pll = std::pair<i64, i64>;
using vi = std::vector<i32>;
using vll = std::vector<i64>;
using vpii = std::vector<pii>;
using vpll = std::vector<pll>;

void solve();

int main()
{
    std::cin.tie(nullptr)->sync_with_stdio(false);
    solve();
}

using distance_group = pii;
using string_group = std::pair<std::string, std::string>;

distance_group get_distance(char a, char b)
{
    auto dis1 = std::abs(a - b);
    auto dis2 = 26 - dis1;
    return {dis1, dis2};
}

string_group generate(const std::string &str, i32 distance)
{
    std::string res1, res2;
    for (auto &item : str)
    {
        res1 += static_cast<char>(item + distance);
        res2 += static_cast<char>(item - distance);
    }
    std::sort(res1.begin(), res1.end());
    std::sort(res2.begin(), res2.end());
    return {res1, res2};
}

void solve()
{
    std::string str1, str2;
    std::cin >> str1 >> str2;
    if (str1.length() != str2.length())
    {
        std::cout << "NO";
        return;
    }
    std::sort(str1.begin(), str1.end());
    bool flag = false;
    for (u64 i = 0; i < str1.length(); ++i)
    {
        auto dis_group = get_distance(str1[i], str2[i]);
        auto str_group_1 = generate(str2, dis_group.first);
        auto str_group_2 = generate(str2, dis_group.second);
        if (str1 == str_group_1.first || str1 == str_group_1.second || str1 == str_group_2.first ||
            str1 == str_group_2.second)
        {
            flag = true;
            break;
        }
    }
    if (flag)
    {
        std::cout << "YES";
        return;
    }
    std::cout << "NO";
}