2017ACM/ICPC广西邀请赛 1005 CS Course

CS Course

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

 

 

题解：这道题目  要注意的是   与   或   异或  的操作    不受前后循序的影响的 

我是使用了6个数组   保存了数组与   或   异或的前缀和   和   后缀    

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <cstring>
#include <math.h>
using namespace std;
#define MAXN 0xffffff
int a[100100];
int qy[100100],hy[100100];//与的前缀  后缀
int qh[100100],hh[100100];//或的前缀   后缀
int qyh[100100],hyh[100100];//异或的
int main()
{
    // & | ^
    int n,q;
    while(~scanf("%d%d",&n,&q))
    {
        scanf("%d",&a[1]);
        qy[1]=qh[1]=qyh[1]=a[1];
        for(int i=2; i<=n; ++i)
        {
            scanf("%d",&a[i]);
            qy[i]=a[i]&qy[i-1];
            qh[i]=a[i]|qh[i-1];
            qyh[i]=a[i]^qyh[i-1];
        }
        hy[n]=hh[n]=hyh[n]=a[n];
        //  printf("%d ",a[n]);
        for(int i=n-1; i>0; --i)
        {
            hy[i]=a[i]&hy[i+1];
            hh[i]=a[i]|hh[i+1];
            hyh[i]=a[i]^hyh[i+1];
        }
      /*  for(int i=1; i<=n; ++i)
        {
            printf("%d ",hh[i]);
        }*/
        while(q--)
        {
            int m;
            scanf("%d",&m);
            // printf("%d %d %d %d %d %d
",qy[m-1],hy[m+1],qh[m-1],hh[m+1],qyh[m-1],hyh[m+1]);
            if(m==1)
            {
                 printf("%d %d %d
",hy[m+1],hh[m+1],hyh[m+1]);
            }
            else if(m==n)
            {
                printf("%d %d %d
",qy[m-1],qh[m-1],qyh[m-1]);
            }
            else
            {
                 printf("%d %d %d
",qy[m-1]&hy[m+1],qh[m-1]|hh[m+1],qyh[m-1]^hyh[m+1]);
            }

        }
    }
    return 0;
}