2017 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,⋯,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 3 1 1 1 1 2 3

 

Sample Output

1 1 0 1 1 0 1 1 0

 

/*
* @Author: Administrator
* @Date:   2017-08-31 17:13:07
* @Last Modified by:   Administrator
* @Last Modified time: 2017-08-31 17:45:05
*/
/*
 题意：给你n个数，然后m次查询，每次给你一个q ,问你除去a[q]之后所有数的 and or xor 

 思路：xor可以直接得出，and 和 or 只需要判断一下每位1的个数

 感悟：上学期我以后学分选完了，结果一看XK 的都没选，这学期课挺多的...今下午没打比赛
    在教室想了一下1005和1004下了课就过来实现了，没想到1A了，看来我还是适合紧张的学习
    环境...
*/
#include <bits/stdc++.h>

#define MAXN 100005
#define LL long long
using namespace std;

int n,m;
LL a[MAXN];
int q;
int vis[33];//统计每位的1的个数
LL res1,res2,res3;//未去除的最后结果
LL pos;//指针
LL cur1,cur2,cur3;//最后结果
LL cnt;
int cnt1[33],cnt2[33];//最后结果的二进制，xor操作可O(1)算出来所以不用保存

void cal(LL x){//统计每位上1的个数
    for(int i=0;i<32;i++){
        if(x%2==1){
            vis[i]++;
        }
        x/=2;
    }
}

void po(LL x,int sw){
    for(int i=0;i<32;i++){
        switch(sw){
            case 1:
                cnt1[i]=x%2;
                break;
            case 2:
                cnt2[i]=x%2;
                break;
        }
        x/=2;
    }
}

inline void init(){
    memset(vis,0,sizeof vis);
    cur1=0;
    cur2=0;
    cur3=0;
}

int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
        for(int i=0;i<n;i++){
            scanf("%lld",&a[i]);
            cal(a[i]);
            if(i==0){
                res1=a[i],res2=a[i],res3=a[i];
            }else{
                res1&=a[i];
                res2|=a[i];
                res3^=a[i];
            }
        }
        //将结果处理成二进制
        po(res1,1);po(res2,2);

        for(int i=0;i<m;i++){
            
            cur1=0;
            cur2=0;
            cur3=0;

            scanf("%d",&q);

            //处理and操作
            pos=a[q-1];
            cnt=1;
            for(int i=0;i<32;i++){
                if(pos%2==0){
                    if(vis[i]==n-1){
                        cur1+=cnt;
                    }
                }else{
                    if(cnt1[i]==1){
                        cur1+=cnt;
                    }
                }
                pos/=2;
                cnt*=2;
            }

            //处理or操作
            pos=a[q-1];
            cnt=1;
            for(int i=0;i<32;i++){
                if(pos%2==1){
                    if(vis[i]!=1){
                        cur2+=cnt;
                    }
                }else{
                    if(cnt2[i]==1){
                        cur2+=cnt;
                    }
                }
                pos/=2;
                cnt*=2;
            }
            //处理xor操作
            cur3=res3^a[q-1];
            printf("%lld %lld %lld
",cur1,cur2,cur3);
        }
    }
    return 0;
}