HDU 6186 CS Course (连续位运算)

最新推荐文章于 2020-05-11 19:38:18 发布

weixin_30341745

最新推荐文章于 2020-05-11 19:38:18 发布

阅读量84

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/wrjlinkkkkkk/p/9552002.html

本文介绍了一道关于位运算查询的算法题目，包括and、or、xor操作，并提供了两种解题思路及代码实现。法一通过保存所有元素的每一位bit的和来判断是否需要更改结果；法二利用位运算的特性，预处理前缀和后缀的and/or结果，快速求解除特定数外的位运算结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

CS Course

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 430 Accepted Submission(s): 222

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 , 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 .

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.

Then n non-negative integers follows in a line, for each i in range[1,n].

After that there are q positive integers in q lines, 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 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）

Recommend

liuyiding

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6186

题意：从连续的and/or/xor中去掉某一个。。

法一：保存所有元素的某一位bit的和，查找ap有没有因为自己的某一位而改变全部的or/and，如果有，就改回来

法二：三个位运算都有交换率律结合律，所以可以保存前缀连续and/or和后缀连续and/or，注except第一个或者最后一个元素的时候要特判一下。

另外，因为异或具有性质：a^b^b = a ，所以可以将直接将1~n的异或再异或要去掉的ap即可

法二代码

#include<cstdio>
#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 100000 + 100;

int n,q;
int a[maxn];
int andl[maxn], andr[maxn], orl[maxn], orr[maxn], xorl[maxn], xorr[maxn];
int main(){

    while(~scanf("%d %d", &n, &q)){
    	//memset(a,-1,sizeof(a));
    	for(int i = 1; i <= n; i++){
    		scanf("%d", &a[i]);
		}
		andl[1] = orl[1] = xorl[1] = a[1];
		andr[n] = orr[n] = a[n];
    	for(int i = 2; i <= n; i++){
    		andl[i] =andl[i-1]&a[i];
			orl[i] = orl[i-1]|a[i];
			xorl[i] = xorl[i-1]^a[i] ;
		}
		for(int i = n-1; i >= 1; i--){
			andr[i] = andr[i+1]&a[i];
			orr[i] = orr[i+1]|a[i];
		}
		for(int i = 0; i < q; i++){
			int pp;
			scanf("%d", &pp);
			if(pp==1)printf("%d %d %d\n", andr[2], orr[2], xorl[n]^a[pp]);
			else if(pp==n)printf("%d %d %d\n", andl[pp-1], orl[pp-1], xorl[n]^a[pp]);
			else printf("%d %d %d\n", andl[pp-1]&andr[pp+1], orl[pp-1]|orr[pp+1], xorl[n]^a[pp]);
		}
		
	}
    return 0;
}

转载于:https://www.cnblogs.com/wrjlinkkkkkk/p/9552002.html