[广西邀请赛

我多久没更新了来着.......?

CS Course

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 282    Accepted Submission(s): 137

Problem Description

Little A has come to college and majored in Computer and Science.

Today he has learned bit-operations in Algorithm Lessons, and he got a problem as homework.

Here is the problem:

You are giving n non-negative integers a1,a2,⋯,an, and some queries.

A query only contains a positive integer p, which means you 
are asked to answer the result of bit-operations (and, or, xor) of all the integers except ap.

 

Input

There are no more than 15 test cases. 

Each test case begins with two positive integers n and p
in a line, indicate the number of positive integers and the number of queries.

2≤n,q≤105

Then n non-negative integers a1,a2,⋯,an follows in a line, 0≤ai≤109 for each i in range[1,n].

After that there are q positive integers p1,p2,⋯,pqin q lines, 1≤pi≤n for each i in range[1,q].

 

Output

For each query p, output three non-negative integers indicates the result of bit-operations(and, or, xor) of all non-negative integers except ap in a line.

 

Sample Input

3 31 1 1123

 

Sample Output

1 1 01 1 01 1 0

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

 

给你n个数a_1 ~ a_n和q次询问，每次询问给你一个数x，输出  所给你的数中，除了a_x数外的剩下数字的 &， | ， ^。

对于异或运算，直接算出所有数字的异或后再异或一次a_x

原理大概就是 a ^ b ^ b = a（这TM叫什么原理）

对于& 和 | 有两种解法

解法一 

先预处理一发所给数列的前缀 &，| 和后缀 &，|，然后直接查询x - 1的前缀&以及x + 1的后缀&，再算一次&就出来了，|也是同理。不过要注意一下初始化的不同

#include <bits/stdc++.h>using namespace std;const int N = 100010;int ma[N];int a1[N], a2[N];int b1[N], b2[N];int c;int main(){    int n, m;    while(scanf("%d%d", &n, &m) == 2){        for(int i = 1; i <= n; i ++){            scanf("%d", &ma[i]);        }        a1[0] = (1 << 31) - 1;/// & 运算的初始值要每一位都为1        b1[0] = 0;        c = 0;        for(int i = 1; i <= n; i ++){            a1[i] = a1[i - 1] & ma[i];            b1[i] = b1[i - 1] | ma[i];            c ^= ma[i];        }        a2[n + 1] = (1 << 31) - 1;        b2[n + 1] = 0;        for(int i = n; i > 0; i --){            a2[i] = a2[i + 1] & ma[i];            b2[i] = b2[i + 1] | ma[i];        }        for(int i = 0; i < m; i ++){            int x;            scanf("%d", &x);            printf("%d %d %d\n", a1[x - 1] & a2[x + 1], b1[x - 1] | b2[x + 1], c ^ ma[x]);        }    }    return 0;}

解法二

开一个cnt数组计数，cnt[i]表示 二进制表示 第 i 位为 1 的数字的个数。

然后预处理一发所有数据的 & 和 |  ，分别记作 A, B

然后我们先看 & 操作

如果A 的二进制第i位为1，那么说明所有数的第i位都为1，那么就无影响