习题8-3(uva-12545)

#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
string s, t;

int main()
{
    
    
	int cnt,kCase = 0;
	cin >> cnt;
	while (cnt--) {
    
    
		cin >> s >> t;
		int t0 = 0, t1 = 0, s0 = 0, s1 = 0, sq = 0, tq0 = 0, tq1 = 0;
		for (int i = 0; i < s.size(); i++) {
    
    
			if (s[i] == '?') {
    
    
				sq++;
				if (t[i] == '0')tq0++;
				else tq1++;
			}
			else if (s[i] != t[i]){
    
    
				if (s[i] == '0') s0++;
				else s1++;

				if (t[i] == '0') t0++;
				else t1++;
			}
		}
		cout << "Case " << ++kCase << ": ";
		//1的个数太多了 1不能减少只能增加
		if (s1 > t1 + tq1) cout << -1 << endl;
		else {
    
    
			int ans = sq;
			/*
				s0 + s1 == t0 + t1
				s0 == t1
				s1 == t0
				现变成下面的形式(不用?)(相对位置)
				111????
				000xxxx
				能不用？交换就不用
				因为0可以变为1,一定能变
			*/
			//交换变成上面的形式
			//先交换
			int tmp = min(s0, s1);
			ans += tmp;
			s0 -= tmp, s1 -= tmp;
			/*
			  s0 == 0 完美
			  s1 == 0 
			*/
			if (s1 == 0) ans += s0;

			ans +=  s1;
			cout << ans << endl;
		}
	}
	return 0;
}