hdu 6186 CS Course 前缀后缀问题

CS Course

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1166    Accepted Submission(s): 560

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

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）  

#include
#include
#include
using namespace std;
#define nsize 100010
int a[nsize];
int frontAnd[nsize], frontOr[nsize], frontXor[nsize];
int behindAnd[nsize], behindOr[nsize], behindXor[nsize];

int main()
{
	int n, m;
	int q;
	while (cin >> n >> m) {
		cin >> a[1];
		frontAnd[1] = frontOr[1] = frontXor[1] = a[1];
		for (int i = 2; i <= n; i++) {
			cin >> a[i];
			frontAnd[i] = a[i] & frontAnd[i - 1];
			frontOr[i] = a[i] | frontOr[i - 1];
			frontXor[i] = a[i] ^ frontXor[i - 1];
		}
		behindAnd[n] = behindOr[n] = behindXor[n] = a[n];
		for (int i = n - 1; i >= 1; i--) {
			behindAnd[i] = behindAnd[i + 1] & a[i];
			behindOr[i] = behindOr[i + 1] | a[i];
			behindXor[i] = behindXor[i + 1] ^ a[i];
		}
		for (int i = 0; i < m; i++) {
			cin >> q;
			if (q == 1)
				printf("%d %d %d\n", behindAnd[q + 1], behindOr[q + 1], behindXor[q + 1]);
			else if (q == n)
				printf("%d %d %d\n", frontAnd[q - 1], frontOr[q - 1], frontXor[q - 1]);
			else
				printf("%d %d %d\n", frontAnd[q - 1] & behindAnd[q + 1], frontOr[q - 1] | behindOr[q + 1], frontXor[q - 1] ^ behindXor[q + 1]);
		}
	}
}