广西 2017 邀请赛 CS Course

题目描述

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.

输入

There are no more than 15 test cases.
Each test case begins with two positive integers n(2 ≤ n ≤ 105) and p(2 ≤ p ≤ 105) in a line, indicate the number of positive integers and the number of queries.
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].

输出

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.

样例输入

3 3
1 1 1
1
2
3

样例输出

1 1 0
1 1 0
1 1 0

题意

一个数n，一个数k，下一行有n个数，再下一行有k个操作。
每一个操作有一个数x，输出，除了x以外的其他的所有的数的且，或，抑或的值。
用4个数组来初始化这个长度为n的序列。
一个存正向的，第k个数之前所有的数的且值，另一个存从第n个数到第n-k个数的且值;
一个存正向的，第k个数之前所有的数的或值，另一个存从第n个数到第n-k个数的或值;
最后还有一个数存数组所有数的抑或值。（因为相同的数抑或两次没有影响。）
然后，读入一个x直接输出。

代码

#include <bits/stdc++.h> //这种思路也是一样的，就是把抑或也跟且、或做了相同的处理。using namespace std;int a[100000],b1[100000],b2[100000],c1[100000],c2[100000],d1[100000],d2[100000];int main(){    int n,p,q,i;    while(scanf("%d %d",&n,&q)!=EOF)    {        for(i=0;i<n;i++)            scanf("%d",&a[i]);        b1[0] = a[0];        c1[0] = a[0];        d1[0] = a[0];        b2[n-1] = a[n-1];        c2[n-1] = a[n-1];        d2[n-1] = a[n-1];        for(i=1;i<n;i++)        {            b1[i] = b1[i-1]&a[i];            c1[i] = c1[i-1]|a[i];            d1[i] = d1[i-1]^a[i];        }        for(i=n-2;i>=0;i--)        {            b2[i] = b2[i+1]&a[i];            c2[i] = c2[i+1]|a[i];            d2[i] = d2[i+1]^a[i];        }        for(i=0;i<q;i++)        {            scanf("%d",&p);            if(p==1)                printf("%d %d %d\n",b2[1],c2[1],d2[1]);            else if(p==n)                printf("%d %d %d\n",b1[n-2],c1[n-2],d1[n-2]);            else                printf("%d %d %d\n",b1[p-2]&b2[p],c1[p-2]|c2[p],d1[p-2]^d2[p]);        }    }    return 0;}