2017ICPC/广西邀请赛1005（水）HDU6186

CS Course

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 593    Accepted Submission(s): 288

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,⋯,pq

in 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

题意  求n个整数（除去） 二进制位运算 和 或 异或的结果

解析   求出前后缀询问时直接前缀后缀计算 

AC代码

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <queue>
#include <vector>
using namespace std;
const int maxn= 1e5 + ;
const int inf = 0x3f3f3f3f;
typedef long long ll;
int n,m;
int sum1[maxn],sum2[maxn],sum3[maxn];
int rsum1[maxn],rsum2[maxn],rsum3[maxn];
int a[maxn];
int main(int argc, char const *argv[])
{
    while(scanf("%d %d",&n,&m)==)
    {
        scanf("%d",&a[]);
        sum1[]=sum2[]=sum3[]=a[];
        for(int i=;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum1[i]=sum1[i-] & a[i];
            sum2[i]=sum2[i-] | a[i];
            sum3[i]=sum3[i-] ^ a[i];
        }
        rsum1[n]=rsum2[n]=rsum3[n]=a[n];
        for(int i=n-;i>=;i--)
        {
            rsum1[i]=rsum1[i+] & a[i];
            rsum2[i]=rsum2[i+] | a[i];
            rsum3[i]=rsum3[i+] ^ a[i];
            //cout<<rsum1[i]<<" "<<rsum2[i]<<" "<<rsum3[i]<<endl;
        }
        int q;
        while(m--)
        {
            scanf("%d",&q);
            if(q==)
                printf("%d %d %d\n",rsum1[],rsum2[],rsum3[]);
            else if(q == n)
                printf("%d %d %d\n",sum1[n-],sum2[n-],sum3[n-]);
            else
                printf("%d %d %d\n",sum1[q-] & rsum1[q+],sum2[q-]|rsum2[q+],sum3[q-]^rsum3[q+]);
        }
    }
    return ;
}