assertThat(7, not(CombinableMatcher.<Integer> either(equalTo(3)).or(equalTo(4))));是什么意思

这是一个用于测试的断言语句，它的意思是：断言 7 不等于 3 或者 4。如果断言成功，则测试通过；否则，测试失败。更具体地说：

- `assertThat` 是 JUnit 提供的一个静态方法，用于进行断言测试。
- `not` 是一个 Matcher，表示取反操作，它的作用是将后面的 Matcher 包装起来，表示其结果的相反值。
- `CombinableMatcher` 是一个 Matcher，表示可以组合多个 Matcher 使用。
- `<Integer>` 是泛型参数，表示匹配的对象类型为 Integer。
- `either` 方法是 `CombinableMatcher` 提供的方法，用于将两个 Matcher 进行逻辑或运算。
- `equalTo` 方法是一个 Matcher，表示等于某个值。
- 因此，`equalTo(3).or(equalTo(4))` 表示匹配值等于 3 或者等于 4。
- 最终，`not(CombinableMatcher.<Integer> either(equalTo(3)).or(equalTo(4)))` 表示匹配值既不等于 3 也不等于 4，即不等于 3 或 4。

因此，整个断言语句的意思是：断言 7 不等于 3 或 4。如果断言成功，则测试通过；否则，测试失败。

相关问题

You have two binary strings � a and � b of length � n. You would like to make all the elements of both strings equal to 0 0. Unfortunately, you can modify the contents of these strings using only the following operation: You choose two indices � l and � r ( 1 ≤ � ≤ � ≤ � 1≤l≤r≤n); For every � i that respects � ≤ � ≤ � l≤i≤r, change � � a i ​ to the opposite. That is, � � : = 1 − � � a i ​ :=1−a i ​ ; For every � i that respects either 1 ≤ � < � 1≤i<l or � < � ≤ � r<i≤n, change � � b i ​ to the opposite. That is, � � : = 1 − � � b i ​ :=1−b i ​ . Your task is to determine if this is possible, and if it is, to find such an appropriate chain of operations. The number of operations should not exceed � + 5 n+5. It can be proven that if such chain of operations exists, one exists with at most � + 5 n+5 operations. Input Each test consists of multiple test cases. The first line contains a single integer � t ( 1 ≤ � ≤ 1 0 5 1≤t≤10 5 ) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer � n ( 2 ≤ � ≤ 2 ⋅ 1 0 5 2≤n≤2⋅10 5 ) — the length of the strings. The second line of each test case contains a binary string � a, consisting only of characters 0 and 1, of length � n. The third line of each test case contains a binary string � b, consisting only of characters 0 and 1, of length � n. It is guaranteed that sum of � n over all test cases doesn't exceed 2 ⋅ 1 0 5 2⋅10 5 . Output For each testcase, print first "YES" if it's possible to make all the elements of both strings equal to 0 0. Otherwise, print "NO". If the answer is "YES", on the next line print a single integer � k ( 0 ≤ � ≤ � + 5 0≤k≤n+5) — the number of operations. Then � k lines follows, each contains two integers � l and � r ( 1 ≤ � ≤ � ≤ � 1≤l≤r≤n) — the description of the operation. If there are several correct answers, print any of them.

这道题目的意思是给定两个长度为n的01字符串a和b，每次操作可以选择一个区间[l, r]，将a中该区间内的元素取反，将b中该区间外的元素取反，问是否能通过有限次操作将a和b都变成全0。

思路：首先考虑每次操作对a和b的影响，发现每次操作不会改变a和b的异同性质，即a和b中相同位置的元素要么都取反，要么都不变。因此我们可以将a和b的不同位置分别记录下来，然后将它们的异同性质分开考虑。对于a和b中相同位置的元素，如果它们不同，那么无论怎么操作都不可能使它们都变成0；如果它们相同，那么我们只需要将它们都变成0即可。对于a和b中不同位置的元素，我们可以将它们分别记录下来，然后我们需要将它们都变成0，那么考虑对于每一个不同位置，我们需要将它所在的位置变成0，同时我们需要保证其他不同位置元素不受影响，因此我们可以将每个不同位置拆分成两个区间[l, r]，分别表示需要将a中[l, r]变成0和需要将b中[l, r]变成0。

具体实现时，我们可以先将a和b中不同的位置分别记录下来，然后对于a和b中相同位置的元素，如果它们都是1，那么我们需要将它们变成0，此时操作数加1；对于a和b中不同位置的元素，我们可以先将每个不同位置拆分成两个区间[l, r]，分别表示需要将a中[l, r]变成0和需要将b中[l, r]变成0。然后我们可以通过贪心的思想，将所有需要变成0的区间按照左端点排序，然后依次进行操作，直到把所有的区间变成0。具体实现时，我们可以将需要变成0的区间看成点，如果两个区间有重叠，那么它们之间就有一条边，我们只需要求出这个图的一个完美匹配即可。

时间复杂度为O(nlogn+n)，其中n为字符串长度。

You are given a string s consisting of the characters 0, 1 and/or ?. Let's call it a pattern. Let's say that the binary string (a string where each character is either 0 or 1) matches the pattern if you can replace each character ? with 0 or 1 (for each character, the choice is independent) so that the strings become equal. For example, 0010 matches ?01?, but 010 doesn't match 1??, ??, or ????. Let's define the cost of the binary string as the minimum number of operations of the form "reverse an arbitrary contiguous substring of the string" required to sort the string in non-descending order. You have to find a binary string with the minimum possible cost among those that match the given pattern. If there are multiple answers, print any of them. Input The first line contains a single integer t (1≤t≤3⋅104 ) — the number of test cases. The first and only line of each test case contains the string s (1≤|s|≤3⋅105 ) consisting of characters 0, 1, and/or ?. The sum of the string lengths over all test cases does not exceed 3⋅105 . Output For each test case, print a binary string with the minimum possible cost among those that match the given pattern. If there are multiple answers, print any of them. Example

这是一道编程题，需要你编写程序来解决。以下是C++的代码实现，可以供参考：

#include <bits/stdc++.h>
using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        string s;
        cin >> s;
        int n = s.size();
        vector<int> q;
        for (int i = 0; i < n; i++) {
            if (s[i] == '?') {
                q.push_back(i);
            }
        }
        if (q.empty()) {
            int cnt = 0;
            for (int i = 1; i < n; i++) {
                if (s[i] != s[i-1]) {
                    cnt++;
                }
            }
            if (cnt == 0) {
                cout << s << endl;
            } else if (cnt == 1) {
                cout << s << endl;
            } else {
                cout << "NO" << endl;
            }
            continue;
        }
        bool ok = false;
        string ans(n, '0');
        for (int i = 0; i < (1<<q.size()); i++) {
            for (int j = 0; j < q.size(); j++) {
                ans[q[j]] = (i>>j)&1 ? '1' : '0';
            }
            int cnt = 0;
            for (int j = 1; j < n; j++) {
                if (ans[j] != ans[j-1]) {
                    cnt++;
                }
            }
            if (cnt == q.size() || cnt == q.size()+1) {
                ok = true;
                break;
            }
        }
        if (ok) {
            cout << ans << endl;
        } else {
            cout << "NO" << endl;
        }
    }
    return 0;
}

该代码先处理出原始字符串中所有的问号的位置，然后枚举所有可能的填充方案，对于每个方案，计算出填充后的字符串的最小排序代价，判断是否符合要求，如果符合，则输出该方案，否则输出NO。