广西 2017 邀请赛 CS Course

xiaoxiao2021-02-28  131

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; }
转载请注明原文地址: https://www.6miu.com/read-17561.html

最新回复(0)