题解 | 字符串合并处理

#include <algorithm>
#include <iostream>
#include<bits/stdc++.h>
#include <vector>
using namespace std;
char change(char c){
    switch(c){
        case '1':c='8';break;
        case '2':c='4';break;
        case '3':c='C';break;
        case '4':c='2';break;
        case '5':c='A';break;
        case '6':c='6';break;
        case '7':c='E';break;
        case '8':c='1';break;
        case '9':c='9';break;
        case 'a':c='5';break;
        case 'A':c='5';break;
        case 'b':c='D';break;
        case 'B':c='D';break;
        case 'c':c='3';break;
        case 'C':c='3';break;
        case 'd':c='B';break;
        case 'D':c='B';break;
        case 'e':c='7';break;
        case 'E':c='7';break;
        case 'f':c='F';break;
        case 'F':c='F';break;
    }
    return c;
}

int main() {
    string s,t;
    cin>>s>>t;
    string u=s.append(t);
    string u1,u2;
    for(int i=0;i<u.size();i++){
        if(i%2==0)u1.push_back(u[i]);
        else u2.push_back(u[i]);
    }
    sort(u1.begin(), u1.end());
    sort(u2.begin(), u2.end());
    for(int i=0;i<u.size();i++){
        if(i%2==0)u[i]=u1[i/2];
        else u[i]=u2[i/2];
        u[i]=change(u[i]);
    }
    cout<<u<<endl;
}
// 64 位输出请用 printf("%lld")

中级偷懒大法之人脑比电脑好算