Ai+ Aj)( 1≤i,j≤n) The xor of an array B is defined as B1 xor B2...xor Bn Input Multiple test cases, the first line contains an integer T(no more than 20), indicating the number of cases. Each test case contains four integers: n, m, z, l A1=0, Ai=(Ai−1∗m+z) ...
MZL loves xor very much.Now he gets an array A.The length of Aisn.He wants to know the xor of all (Ai+Aj)(1≤i,j≤n) The xor of an array BisdefinedasB1 xor B2...xor Bn Input Multiple test cases, the first line contains an integer T(no more than20), indicating the number ...
MZL loves xor very much.Now he gets an array A.The length of A is n.He wants to know the xor of all (Ai+Aj)(1≤i,j≤n)The xor of an array B is defined as B1 xor B2...xor Bn Input Multiple test cases, the first line contains an integer T(no more than 20), indicating ...
The xor of an array BisdefinedasB1 xor B2...xor Bn 1. 2. Input Multiple test cases, the first line contains an integer T(no more than20), indicating the number of cases. Each testcasecontains four integers:n,m,z,l A1=0,Ai=(Ai−1∗m+z) mod l1≤m,z,l≤5∗105,n=5∗...
这题真的是对mask的理解升华了 还有一个要注意的是xor的意义。我一直觉得做一个operation需要两个数,要save一个之前的,再apply到当前的数。 但是这题不用。...
然后整数类型最多31位,所以从高往低,用&操作来截取高位,然后通过上面这个知识点来判断当前位是否存在两个数异或使得当前位为1,并且高位的那些还是保持原来的。 classSolution{public:intfindMaximumXOR(vector<int>&nums){intmask=0,maximum=0;for(inti=31;i>=0;i--){mask|=1<<i;unordered_set<int>prefixs...
XOR of Sum of every possible pair of an array in C - In this problem, we are given an array of n elements. Our task is to generate a sequence of size n*n whose elements are the sum of a pair of all elements of A with itself. And print the xor elements o
demons_paw's blog Given an array of sizenn. How can we find sum of XOR sum of all subsets in better thanO(2n)O(2n)? For example considerarray=[1,4,5]array=[1,4,5] Answer = 1 + 4 + 5 + 1^4 + 1^5 + 4^5 + 1^4^5Answer = 1 + 4 + 5 + 1^4 + 1^5 + 4^5 ...
public int findMaximumXOR(int[] nums) { /* trie tree: root is the largest bit * 32 bits */ TrieNode root = new TrieNode(); for(int num : nums) { TrieNode node = root; for(int i = 31; i >= 0; i--) { int bit = (num >> i) & 1; ...
LeetCode[421] Maximum XOR of Two Numbers in an Array Given a non-emptyarrayof numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231. Find the maximum result of ai XOR aj, where 0 ≤ i, j < n. Could you do this in O(n) runtime?