#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;
}
习题8-3(uva-12545)
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Origin blog.csdn.net/seanbill/article/details/116426705
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