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[];
 int qy[],hy[];//与的前缀  后缀
 int qh[],hh[];//或的前缀   后缀
 int qyh[],hyh[];//异或的
 int main()
 {
     // & | ^
     int n,q;
     while(~scanf("%d%d",&n,&q))
     {
         scanf("%d",&a[]);
         qy[]=qh[]=qyh[]=a[];
         for(int i=; i<=n; ++i)
         {
             scanf("%d",&a[i]);
             qy[i]=a[i]&qy[i-];
             qh[i]=a[i]|qh[i-];
             qyh[i]=a[i]^qyh[i-];
         }
         hy[n]=hh[n]=hyh[n]=a[n];
         //  printf("%d ",a[n]);
         for(int i=n-; i>; --i)
         {
             hy[i]=a[i]&hy[i+];
             hh[i]=a[i]|hh[i+];
             hyh[i]=a[i]^hyh[i+];
         }
       /*  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\n",qy[m-1],hy[m+1],qh[m-1],hh[m+1],qyh[m-1],hyh[m+1]);
             if(m==)
             {
                  printf("%d %d %d\n",hy[m+],hh[m+],hyh[m+]);
             }
             else if(m==n)
             {
                 printf("%d %d %d\n",qy[m-],qh[m-],qyh[m-]);
             }
             else
             {
                  printf("%d %d %d\n",qy[m-]&hy[m+],qh[m-]|hh[m+],qyh[m-]^hyh[m+]);
             }

         }
     }
     return ;
 }