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,⋯,pq
in 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 3
1 1 1
1
2
3
Sample Output
1 1 0
1 1 0
1 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
