点击打开链接

CS Course

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1699 Accepted Submission(s): 726

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）

Recommend

liuyiding | We have carefully selected several similar problems for you: 6275 6274 6273 6272 6271

题意：

给定n个数，然后是q次询问，每次询问为去掉第i个数，输出剩下n-1个数与、或、异或的结果

解法1：

前缀数组和后缀数组

用p1[i]表示 [1,i] 所有数与的结果，b1[i]表示 [i,n] 所有数与的结果

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int a[maxn];
int p1[maxn],b1[maxn];//and
int p2[maxn],b2[maxn];//or
int p3[maxn],b3[maxn];//xor
int main()
{ios::sync_with_stdio(0);int n,q,p;while(cin>>n>>q){cin>>a[1];p1[1]=p2[1]=p3[1]=a[1];for(int i=2;i<=n;i++){cin>>a[i];p1[i]=p1[i-1]&a[i];p2[i]=p2[i-1]|a[i];p3[i]=p3[i-1]^a[i];}b1[n]=b2[n]=b3[n]=a[n];for(int i=n-1;i>=1;i--){b1[i]=b1[i+1]&a[i];b2[i]=b2[i+1]|a[i];b3[i]=b3[i+1]^a[i];}while(q--){cin>>p;if(p==1)printf("%d %d %d\n",b1[p+1],b2[p+1],b3[p+1]);else if(p==n)printf("%d %d %d\n",p1[p-1],p2[p-1],p3[p-1]);elseprintf("%d %d %d\n",p1[p-1]&b1[p+1],p2[p-1]|b2[p+1],p3[p-1]^b3[p+1]);}}return 0;
}

解法2：

用num数组记录n个数转为二进制后，每一位上1的个数之和，

当去掉一个数后，更新num数组，然后遍历num数组

如果num[i]==n-1，那么这一位与的结果为1，否则为0（与运算：全1为1，有0为0）

如果num[i]>=1，那么这一位或的结果为1，否则为0，（或运算：全0为0，有1为1）

对于异或运算：

先让n个数异或的结果为XOR，然后XOR和去掉的数字异或，结果就是n-1个数字异或的结果（异或运算的逆运算）

#include<bits/stdc++.h>
#include<bitset>
using namespace std;
const int maxn=1e5+5;
int a[maxn];
int num[32];int main()
{ios::sync_with_stdio(0);int n,q,p;while(cin>>n>>q){int XOR=0;fill(num,num+32,0);for(int i=1;i<=n;i++){cin>>a[i];XOR^=a[i];bitset<32>bit(a[i]);for(int j=0;j<32;j++)if(bit[j]==1)num[j]++;}while(q--){cin>>p;bitset<32>bit(a[p]);bitset<32>AND;//n-1个数字与的结果bitset<32>OR;//n-1个数字或的结果for(int j=0;j<32;j++){if(num[j]-bit[j]==(n-1))//这一位有n-1个1，与的结果为1AND.set(j);//把第j位置1elseAND.reset(j);//把第j位置0if(num[j]-bit[j]>=1)//这一位上有1，或的结果为1OR.set(j);elseOR.reset(j);}printf("%d %d %d\n",AND,OR,XOR^a[p]);}}return 0;
}

HDU 6186 CS Course相关推荐

HDU - 6186 CS Course(维护前缀+后缀)
题目链接:点击查看题目大意:给出n个数以及m个查询,每个查询包括一个数字,要求输出除了该数字之外的位运算的"或和","与和"和"异或和" ...
HDU 6186 CS Course（线段树区间操作）
Little A has come to college and majored in Computer and Science. Today he has learned bit-operation ...
【HDU - 6186】CS Course（按位与，按位或，按位异或的区间统计，二进制拆位）
题干: Little A has come to college and majored in Computer and Science. Today he has learned bit-opera ...
HDU 6186 2017广西邀请赛：CS Course
题意: n个数,m次查询,每次给出一个p,求出除了第p个数以外其它所有数的且和,或和,异或和统计下每个二进制位1的数量,例如z[5] = x表示有x个数第5个二进制位为1 之后每次查询只要check ...
CS Course HDU - 6186
Little A has come to college and majored in Computer and Science. Today he has learned bit-operation ...
2017ACM/ICPC广西邀请赛
2017ACM/ICPC广西邀请赛(感谢广西大学) 题号题目考点难度 A A Math Problem 数论签到题 B Color it C Counting Stars D Covering ...
hdu 1174:爆头（计算几何，三维叉积求点到线的距离）
爆头 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submiss ...
poj 2096 , zoj 3329 , hdu 4035 —— 期望DP
题目:http://poj.org/problem?id=2096 题目好长...意思就是每次出现 x 和 y,问期望几次 x 集齐 n 种,y 集齐 s 种: 所以设 f[i][j] 表示已经有几种 ...
HDU 4685. Prince and Princess
链接 http://acm.hdu.edu.cn/showproblem.php?pid=4685 题意 nnn 个王子,mmm 个公主,每个王子可以娶他喜欢的公主中的一个,每个公主只能嫁个一个王子, ...

HDU 6186 CS Course