【2017广西邀请赛】hdu 6186 CS Course

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

 

题意：

求n个数，q个询问，求除了第p个数的&，|，^。

思路：

异或的话直接求出所有的数的异或，然后异或询问值。

与，或的话统计所有数每个位上的值。如果剩余数在当前位上的值大于0，则或值加上这个位的值。

如果剩余数早当前位上的值等于n-1，则与值加上这个位上的值。

////  main.cpp//  1005////  Created by zc on 2017/9/6.//  Copyright © 2017年 zc. All rights reserved.//#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define ll long longusing namespace std;const int N=110000;int n,q,b[N][40],c[40],p;ll a[N];int main(int argc, const char * argv[]) {    while(~scanf("%d%d",&n,&q))    {        ll ans3=0;        memset(b,0,sizeof(b));        memset(c,0,sizeof(c));        for(int i=1;i<=n;i++)        {            scanf("%lld",&a[i]);            ans3^=a[i];            for(int j=0;j<33;j++)            {                if(a[i]&(1LL<<j))   b[i][j]=1;                c[j]+=b[i][j];            }        }        for(int i=0;i<q;i++)        {            scanf("%d",&p);            ll ans1=0,ans2=0;            for(int j=0;j<33;j++)            {                if(c[j]-b[p][j]==n-1)   ans1+=(1LL<<j);                if(c[j]-b[p][j]>0)  ans2+=(1LL<<j);            }            printf("%lld %lld %lld\n",ans1,ans2,ans3^a[p]);        }    }}