题解 | #字符串合并处理#

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <math.h>
using namespace std;

string strMerge(string s1, string s2)
{
	string s = s1 + s2;
	vector<char> odd;
	vector<char> even;
	size_t len = s.length();

	for (size_t i=0; i<len; i++)
	{
		if (i%2 == 0)
		{
			odd.push_back(s[i]);
		}
		else
		{
			even.push_back(s[i]);
		}
	}

	sort(odd.begin(), odd.end());
	sort(even.begin(), even.end());

	s.clear();
	int p = 0;
	int q = 0;
	for (size_t i=0; i<len; i++)
	{
		if (i%2 == 0)
		{
			s.push_back(odd[p++]);
		}
		else
		{
			s.push_back(even[q++]);
		}
	}

	return s;
}

char IntReverse(char ch)
{
	string s;
	int val;

	if (ch >= '0' && ch <= '9')
	{
		val = ch - '0';
	}
	else if (ch >= 'a' && ch <= 'f')
	{
		val = ch - 'a' + 10;
	}
	else if (ch >= 'A' && ch <= 'F')
	{
		val = ch - 'A' + 10;
	}
	else
	{
		return ch;
	}

	while (val > 0)
	{
		char ch = (val % 2 == 1) ? '1' : '0';
		s.push_back(ch);
		val /= 2;
	}

	while (s.length() < 4)
	{
		s.push_back('0');
	}

	int i = 0;
	int sum = 0;
	for (auto it = s.rbegin(); it<s.rend(); it++)
	{
		sum += (*it-'0') * int(pow(2, i));
		i++;
	}

	if (sum < 10)
	{
		return '0' + sum;
	}

	return 'A' + sum - 10;
}

int main() {
    string s1, s2;

    cin >> s1 >> s2;
    string s = strMerge(s1, s2);

    for (char ch : s)
    {
        cout << IntReverse(ch);
    }

    cout << endl;
    return 0;
}
// 64 位输出请用 printf("%lld")