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  • Xor

    2019-05-29 09:18:16
    // This file is part of ... // and the book “The Elements of Computing Systems” // by Nisan and Schocken, MIT Press. // File name: projects/01/Xor.hdl /** Exclusive-or gate: out = not...

    CHIP Xor {
    IN a, b;
    OUT out;

    PARTS:
    // Put your code here:
    

    Not(in=a,out=nota);
    Not(in=b,out=notb);
    And(a=nota,b=b,out=w1);
    And(a=a,b=notb,out=w2);
    Or(a=w2,b=w1,out=out);
    }

    展开全文
  • XOR

    千次阅读 2018-06-08 09:20:07
    https://www.lijinma.com/blog/2014/05/29/amazing-xor/
    展开全文
  • xor

    2017-09-20 15:38:41
    给出n个数字a_1,…a_n,问最多有多少个不重叠的非空区间,使得每个区间内数字的xor都等于0,即得出最大的k,使得存在k个区间(l[i],r[i]),满足1<=l[i]<=r[i]<=n,(1<=i<=k),r...

    给出n个数字a_1,…a_n,问最多有多少个不重叠的非空区间,使得每个区间内数字的xor都等于0,即得出最大的k,使得存在k个区间(l[i],r[i]),满足1<=l[i]<=r[i]<=n,(1<=i<=k),r[i] < r[i+1],(1<=i<=k),且a[l[i]] xor a[l[i]+1] xor … xor a[r[i]] = 0,(1<=i<=k)
    输入描述:第一行一个整数n,第二行n个整数a_1,a_2…a_n,对于30%的数据,n<=20,对于100%的数,n<=100000,a_i<=100000;
    输出描述:一个整数表示最多的区间个数

    import java.util.Scanner;
    
    public class Kqujian {
        static int maxn = 1000000;
        static int[] a = new int[maxn];
        static int[] b = new int[maxn];
        static int[] c = new int[maxn];
        public static void main(String[] args) {
            Scanner in = new Scanner(System.in);
            int n = in.nextInt();
            while(in.hasNext()){
                int ans = 0;
                int pos = -1;
                for (int i = 0; i < n; i++) {
                    a[i] = in.nextInt();
                    if(a[i] == 0){
                        ans++;
                        pos = i+1;
                    }
                    b[a[i]]++;
                    if(b[a[i]] != 0 && b[a[i]] % 2 == 0){
                        int temp = c[a[i]];
                        if(temp < pos)
                            continue;
                        int temp2 = 0;
                        for (int j = temp + 1; j < i; j++) {
                            temp2 = temp2 ^ a[j];
                        }
                        if(temp2 == 0){
                            ans++;
                            pos = i+1;
                        }
    
                    }else
                        c[a[i]] = i;
                }
                System.out.println(ans);
            }
        }
    }
    
    展开全文
  • xor_file 另一个xor程序用于xor文件。
  • XOR issue

    2020-11-24 18:16:51
    m using the xor and times functions within this library. However, precision seems to be an issue. <p>With this library: -3750763034362895579 XOR 52 = -3750763034362895400 <p>Correct answer: -...
  • TheXOR

    2020-12-02 22:21:00
    <div><p>.gitmodules for my repository (TheXOR) has an invalid url, it should be corrected as: <p>url = https://github.com/The-XOR/RackPlugins</p> <p>the same for the .json file in manifest folder ...
  • xor helper

    2020-12-26 18:27:39
    <div><p>What do you think about xor helper?</p><p>该提问来源于开源项目:jmurphyau/ember-truth-helpers</p></div>
  • 用一个xor表示多个xorProblem statement: 问题陈述: You are given an array A[] of size N. Now, we call the value of an array the bit-wise XOR of all elements it contains. For example, the value of the...

    用一个xor表示多个xor

    Problem statement:

    问题陈述:

    You are given an array A[] of size N. Now, we call the value of an array the bit-wise XOR of all elements it contains. For example, the value of the array [1,2, 3] = 1^2^3. Now, your task is to find the bit-wise XOR of the values of all sub-arrays of array A.

    给您一个大小为N的数组A [] 。 现在,我们将数组的值称为它包含的所有元素的按位XOR。 例如, 数组[1,2,3]的值= 1 ^ 2 ^ 3 。 现在,您的任务是找到数组A的所有子数组的值的按位XOR。

    Example:

    例:

        Input array:
        1, 2, 3, 4
        Subarrays and their XOR values
        Subarray of length 1:
        1 //value of the subarray: 1
        2 //value of the subarray: 2
        3 //value of the subarray: 3
        4 //value of the subarray: 4
    
        Subarray of length 2:
        1, 2 //value of this subarray:1^2 
        2, 3 //value of this subarray:2^3
        3, 4 //value of this subarray:3^4
        Subarray of length 3
        1, 2, 3 //value of this subarray:1^2^3
        2, 3, 4 // value of this subarray:2^3^4
        Subarray of length 4
        1, 2, 3, 4 //value of this subarray:1^2^3^4
    
        So
        Bitwise XOR of all the values are
        1^2^3^4^(1^2)^(2^3)^(3^4)^(1^2^3)^(2^3^4)^(1^2^3^4)
        =0
    
    

    Solution:

    解:

    One naïve approach can be of course to compute values ( XORing elements) for all subarrays. This is going to be real tedious as we need to find out all subarrays.

    一种简单的方法当然是为所有子数组计算值(XORing元素)。 这将非常繁琐,因为我们需要找出所有子数组。

    Effective approach can be the XOR nature. We know XORing an element twice results 0.

    有效的方法可以是XOR性质。 我们知道对元素进行XOR两次会得到0。

    i^i=0

    i ^ i = 0

    Let's revise our example:

    让我们修改示例:

        For input array:
        1, 2, 3, 4
        Sub arrays are:
        {1}
        {2}
        {3}
        {4}
        {1, 2}
        {2, 3}
        {3, 4}
        {1, 2, 3}
        {2, 3, 4}
        {1, 2, 3, 4}
    
        This is all possible sub arrays
        A prompt observation leads to the fact that every element 
        actually occurred even number of times while XORing all values
        1^2^3^4^(1^2)^(2^3)^(3^4)^(1^2^3)^(2^3^4)^(1^2^3^4)
    
        1 has appeared 4 times
        2 has appeared 6 times
        3 has appeared 6 times
        4 has appeared 4 times
        So the final result is actually 0
    
        Now take any even length array
        You will find all elements appearing even no of times 
        if you expand the XORing sequence like the previous one.
        So,
        For even length array the output will be always 0, 
        no matter what the elements are.
    
        Now what about odd lengths.
        Let's quickly revise an example
        The array be:
        1, 2, 3, 4, 5
        So,
        Subarrays are:
        {1}
        {2}
        {3}
        {4}
        {1, 2}
        {2, 3}
        {3, 4}
        {4, 5}
        {1, 2, 3}
        {2, 3, 4}
        {3, 4, 5}
        {1, 2, 3, 4}
        {2, 3, 4, 5}
        {1, 2, 3, 4, 5}
        So the expanded result sequence would be:
    
        1^2^3^4^5^(1^2)^(2^3)^(3^4)^(4^5)^(1^2^3) ^ 
        (2^3^4) ^ (3^4^5) ^ (1^2^3^4) ^(2^3^4^5) ^ (1^2^3 ^4^5)
        
        So, 1 appears 5 times //resulting 1
        2 appears 8 times //resulting 0
        3 appears 9 times //resulting 3
        4 appears 8 times //resulting 0
        5 appears 5 times //resulting 5
        Output will be : 1^3^5
    
    

    Thus for an odd length array it's found that:

    因此,对于奇数长度的数组,发现:

    Result will be XORing of all even indexed elements (0-based indexing)

    结果将是对所有偶数索引元素进行异或(基于0的索引)

    You can check the argue by doing few more examples your own.

    您可以通过自己做一些其他示例来检查争论。

    So the entire problem actually reduced to a simple concept-

    因此,整个问题实际上简化为一个简单的概念-

    1. Even length array: output 0

      偶数长度数组:输出0

    2. Odd length array: XORing all even-indexed elements

      奇数长度数组:对所有偶数索引元素进行异或

    No need to compute all the sub arrays at all!!

    根本不需要计算所有子数组!!

    C++ implementation:

    C ++实现:

    #include <bits/stdc++.h>
    using namespace std;
    
    int gameOfXor(vector<int> a,int n){
    	if(a.size()%2==0) //for even length array
    		return 0;
    		
    	int sum=a[0];
    	// for odd length array
    	for(int it=2;it<a.size();it=it+2) 
    		sum=sum^a[it];
    	return sum;    
    }
    
    int main(){
    	int n,item;
    
    	cout<<"Enter number of elements\n";
    	scanf("%d",&n);
    	
    	vector<int> a;
    	
    	cout<<"Enter array elements\n";
    	for(int j=0;j<n;j++){
    		scanf("%d",&item);
    		a.push_back(item);
    	}
    	
    	cout<<"Output is:\n";
    	cout<<gameOfXor(a,n)<<endl;
    
    	return 0;
    }
    
    

    Output

    输出量

    First run:
    Enter number of elements
    4
    Enter array elements
    1 2 3 4
    Output is:
    0
    
    Second run:
    Enter number of elements
    5
    Enter array elements
    1 2 3 4 5
    Output is:
    7
    
    
    

    翻译自: https://www.includehelp.com/icp/game-of-xor.aspx

    用一个xor表示多个xor

    展开全文
  • XOR加密

    2020-06-29 14:57:02
    本文介绍一种简单高效、非常安全的加密方法:XOR 加密。 一、 XOR 运算 逻辑运算之中,除了AND和OR,还有一种XOR运算,中文称为"异或运算"。 它的定义是:两个值相同时,返回false,否则返回true。也就是说,...
  • mkdir partner-xor/build && cd partner-xor/build cmake -DCMAKE_INSTALL_PREFIX: < installation> .. make make install (可选)在执行make install之前,您还可以使用make test执行make test 。 请注意,这...
  • XOR 加密

    2019-10-07 08:21:45
    XOR运算 XOR运算,中文称为“异或运算”。 它的定义是:两个值相同时,返回false,否则返回true。也就是说,XOR可以用来判断两个值是否不同。 对应的真值表如下: 输入 A B输出A XOR B 0 0 ...
  • hdu3949 XOR xor高斯消元

    2019-10-06 16:23:56
    XOR Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1364Accepted Submission(s): 402 Problem Description XOR is a kind of bit operato...
  • XOR神经网络-源码

    2021-02-17 14:44:18
    XOR神经网络
  • XOR Clique

    2019-05-16 21:37:46
    异或,英文为exclusive OR,缩写成xor 异或(xor)是一个数学运算符。它应用于逻辑运算。异或的数学符号为“⊕”,计算机符号为“xor”。其运算法则为: a⊕b = (¬a ∧ b) ∨ (a ∧¬b) 异或也叫半加运算,其...
  • Xor

    2017-08-09 19:50:00
    Xor路 (xor.pas/c/cpp)128MB1s 给定一棵有N个点和N-1条边的树,请你求出树中的最长路径,以及总共有多少条最长路径。 这里路径长度是用xor定义的,即若经过的边的权值为a1, a2, a3,...,an,则这条路径的总权值为 ...
  • Xor Path

    2020-02-26 19:15:14
    题目链接:Xor Path 显然对于一个点,如果有偶数条路径通过这个点,那么贡献为0。 所以统计每个点有多少条路径通过这个点即可。 AC代码: #pragma GCC optimize("-Ofast","-funroll-all-loops") #include<bits/...
  • XOR ??

    2021-01-12 03:36:04
    je veux effectuer un Xor entre une image et les frames d'une vidéo j'ai utilisé img = grabber.grab(); pour extraire les frames j'ai pas compris comment effectuer l'xor </p><p>...
  • An XOR example

    2020-12-28 04:42:37
    <p>I have created an XOR example. This could be useful for debugging. Surprisingly, it still does not learn the XOR function. In Pybrain, I managed to get it learn in 20 iterations. <p>All the best,...
  • Xor Sum

    2020-01-16 18:49:40
    题目:Xor Sum 题解:这题运用的是01字典树。是一个模板题。 #include <bits/stdc++.h> using namespace std; const int N = 1e5+10; typedef long long ll; int a[N]; int cnt; int tree[N<<4][2]; ...
  • XOR Permutations

    2019-05-06 00:13:40
    A bitwise XOR takes two bit patterns of equal length and performs the logical exclusive OR operation on each pair of corresponding bits. The result in each position is11if both bits are different (o.....
  • XOR运算

    2020-01-12 11:00:31
    Xor 按位异或 符号在编程语言中通常是 ^ 数学符号通常用⊕表示 X = 0101B Y = 1011B X Y X⊕B 1 0 1 1 1 0 0 0 0 0 1 1 X^Y = 1110B 性质 1. 0 ^ 0 = 0 2.a ^ a = 0 3.0 ^ 1 ^ 2 … ^ n的性质 先...
  • XOR 加密简介

    2020-10-28 16:31:07
    本文介绍一种简单高效、非常安全的加密方法:XOR 加密。 一、 XOR 运算 逻辑运算之中,除了AND和OR,还有一种XOR运算,中文称为"异或运算"。 它的定义是:两个值相同时,返回false,否则返回true。也就是说,...
  • xor b]\sum_{1\leq b \leq a \leq n}[\gcd(a,b)=a~xor~b]∑1≤b≤a≤n​[gcd(a,b)=a xor b]。 题解 有一个很重要的性质:当满足 gcd⁡(a,b)=a xor b\gcd(a,b)=a~xor~...
  • XOR问题

    千次阅读 2019-04-18 20:05:23
    在Rosenblatt单层感知机中已经对异或问题做了介绍,并论证说明了只适用... 异或(XOR)问题可以看做是单位正方形的四个角,响应的输入模式为(0,0),(0,1),(1,1),(1,0)。第一个和第三个模式属于类0,即 和...
  • xor序列

    2019-07-06 15:44:30
    xor序列 题意:给出n个数,问能否将x跟任意的这n个数异或和为y; 思路:线性基。x^z=y ==>x^y=z 其实就是线性基的插入以及判断z是否在线性基里。 #include<bits/stdc++.h> using namespace std; long ...
  • 题目链接   输入整数n(1≤n≤30000000),有多少对整数(a,b)满足:1≤b≤a≤n,且gcd(a,b)=a XORb。例如n=7时,有4对:(3,2), (5,4), (6,4),...不过xor的好处是:a xor b = c,则a xor c = b,所以可以枚举a和c...

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