广西2017邀请赛 Problem E CS Course

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最新推荐文章于 2019-11-07 12:32:25 发布

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分类专栏： # ACM/ICPC题解

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本文介绍了一种基于位运算的算法题解决方案，该问题要求计算一组整数中除指定数外其余数的按位与、或、异或结果。通过统计每位上0的数量并结合查询数，快速得出答案。

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问题 E: CS Course

时间限制: 2 Sec 内存限制: 512 MB

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

样例输出

1 1 0


1 1 0


1 1 0

题目大意：

n个数 q次询问每次询问第pi个数字求除去询问的这个数外其他全部数字的and、or、xor

分析：

分别求出每个二进制位的有多少个0 然后判断去除询问的数字外剩下的每个二进制位的and、or、转为十进制

关于xor 最后在xor一遍这个数即可

AC代码：

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include<list>
#include <bitset>
#include <climits>
#include <algorithm>
#define gcd(a,b) __gcd(a,b)
#define FIN freopen("input.txt","r",stdin)
#define FOUT    freopen("output.txt","w",stdout)
typedef long long LL;
const LL mod=1e9+7;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
using namespace std;
int a[1000005];
int count0[1000005];
int main (){
    int n,p;
    while (scanf ("%d%d",&n,&p)!=EOF){
        memset(a,0,sizeof(a));
        memset(count0,0,sizeof(count0));
        int ans3=0;
        int maxn=0;
        for (int i=0;i<n;i++){
            scanf ("%d",&a[i]);
            ans3^=a[i];
            int tt=a[i];
            int len=0;
            while (tt){
                if(tt%2) count0[len]++;
                tt>>=1;len++;
                if (maxn<len) maxn=len;
            }   
        }
        for (int i=0;i<maxn;i++) count0[i]=n-count0[i];
        int temp;
        for (int i=0;i<p;i++){
            scanf ("%d",&temp);
            int t=a[temp-1];
            int ans1=0,ans2=0;
            for (int j=0;j<maxn;j++){
                if (count0[j]==0||(count0[j]==1&&t%2==0)) ans1+=1<<j,ans2+=1<<j; 
                else if ((count0[j]==n-1&&t%2==0&&n!=2)||
				(count0[j]==1&&t%2!=0&&n!=2)||
				(count0[j]>1&&count0[j]<n-1))
                ans2+=1<<j;
                t>>=1;
            }
            printf ("%d %d %d\n",ans1,ans2,ans3^a[temp-1]);
        }
          
    }
    return 0;
}