HDU.6186.CSCource.(前缀数组和后缀数组)

　　明天后天是南昌赛了嘤嘤嘤，这几天就先不更新每日题目了，以后补题嘤嘤嘤。

今天和队友做了一套2017年广西邀请赛，5个题还是有点膨胀......

　　好了，先来说一下有意思的题目吧......

CS Course

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3450    Accepted Submission(s): 1287

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）

 

Recommend

liuyiding   |   We have carefully selected several similar problems for you:   6543  6542  6541  6540  6539 

 

本题思路：我一开始上来就一发暴力，结果超时了，后来和队友讨论了一下，得出 ^ 可以逆着运算，也就是如果x ^ y = z，那么y 就可以用 z ^ x 得到，&和 | 我们打算将所有位置出现0和1的次数统计和，然后删除数字的时候删除对应数字上的0和1的个数，判断&和|即可。后来没有实现这个，而是用后缀数组和前缀数组实现，接着进行相应的位运算就行了，第一次做到这类型的思维题，所以今天先写一发博客记录。

参考代码：

#include <cstdio>
using namespace std;

const int maxn = 100000 + 5;
int c[maxn], _and1[maxn], _and2[maxn], _or1[maxn], _or2[maxn], _xor1[maxn], _xor2[maxn];

int main() {
    int n, p, q, ans1, ans2, ans3;
    while(~scanf("%d %d", &n, &p)) {
        for(int i = 1; i <=n; i ++) {
            scanf("%d", &c[i]);
        }
        _and1[1] = _or1[1] = _xor1[1] = c[1];
        _and2[n] = _or2[n] = _xor2[n] = c[n];
        for(int i = 2; i <= n; i ++) {
            _and1[i] = _and1[i - 1] & c[i];
            _or1[i] = _or1[i - 1] | c[i];
            _xor1[i] = _xor1[i - 1] ^ c[i];
        }
        for(int i = n - 1; i >= 1; i --) {
            _and2[i] = _and2[i + 1] & c[i];
            _or2[i] = _or2[i + 1] | c[i];
            _xor2[i] = _xor2[i + 1] ^ c[i];
        }
        
        for(int i = 0; i < p; i ++) {
            scanf("%d", &q);
            if(q == 1) {
                ans1 = _and2[2];
                ans2 = _or2[2];
                ans3 = _xor2[2];            
            } else if(q == n) {
                ans1 = _and1[n - 1];
                ans2 = _or1[n - 1];
                ans3 = _xor2[n - 1];
            } else {
                ans1 = _and1[q - 1] & _and2[q + 1];
                ans2 = _or1[q - 1] | _or2[q + 1];
                ans3 = _xor1[q - 1] ^ _xor2[q + 1];
            }
            printf("%d %d %d\n", ans1, ans2, ans3);
        }
    }
    return 0;
}