取反 Single Number

本文介绍了一个利用异或运算解决数组中单数元素问题的算法,通过不断异或相邻元素,最终得到结果为唯一出现一次的元素。

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思想:

1^1 == 0 is true.


class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int i,result=nums[0];
        for(int i=1;i<nums.size();i++) {
            result = result^nums[i];
        }
        return result;
    }
};


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.
06-11
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