题解 | 字符串合并处理

#include <iostream>
#include <string>
#include <bitset>
#include <algorithm>
#include <cctype>
#include <vector>

using namespace std;

const char hex_table[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};

char adjust(const char& c) {
    int n;
    if(isdigit(c)) { //c是'0'-'9'
        n = c - '0';
    } else if(islower(c)) { //c是小写字母
        n = c - 'a' + 10;
    } else if(isupper(c)) { //c是大写字母
        n = c - 'A' + 10;
    }
    if(n > 15) return c; //傻逼题目 说好的只输入a-f 测试用例还输入别的
    bitset<4> b(n);
    string str = b.to_string();
    reverse(str.begin(), str.end());
    bitset<4> newb(str);
    unsigned long ul = newb.to_ulong();
    return hex_table[ul];
}



int main() {
    string s, t, u;
    cin >> s >> t;
    u = s + t;
    // 分离奇偶索引的元素
    vector<char> evenChars; // 偶数索引
    vector<char> oddChars;  // 奇数索引

    for (size_t i = 0; i < u.size(); ++i) {
        if (i % 2 == 0) {
            evenChars.push_back(u[i]);
        } else {
            oddChars.push_back(u[i]);
        }
    }

    // 分别排序
    sort(evenChars.begin(), evenChars.end());
    sort(oddChars.begin(), oddChars.end());

    // 合并结果
    size_t evenIdx = 0, oddIdx = 0;
    for (size_t i = 0; i < u.size(); ++i) {
        if (i % 2 == 0) {
            u[i] = evenChars[evenIdx++];
        } else {
            u[i] = oddChars[oddIdx++];
        }
    }
    string ans;
    for(const char& c: u) {
        ans += adjust(c);
    }
    cout << ans << endl;
    return 0;
}