HDU6186CS Course前缀和

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#ACM #前缀和

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本文提供HDU 6186CSCourse的解题思路与AC代码，该题涉及位运算及查询处理。通过对输入序列的前缀和后缀位运算，高效解答关于特定元素排除后的位与、位或及位异或问题。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6186

6186 CS Course

Time Limit: 4000/2000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 466
Accepted Submission(s): 232

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 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≤10^5

Then n non-negative integers a1,a2,⋯,an follows in a line, 0≤ai≤10
^9 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：给一个序列a，求除去pk个的与，或，异或

思路：分别求前缀数组和后缀数组，在pi的两边分别进行运算，再将这两个结果进行运算即可。

附上ac代码

#include<stdio.h>
using namespace std;
const int maxn = 100005;
int a[maxn];
int x1[maxn], y1[maxn], z1[maxn];
int x2[maxn], y2[maxn], z2[maxn];
int main(){
     int n, m;
     while(scanf("%d%d", &n, &m) != EOF){
          for(int i = 1; i <= n; i++)
               scanf("%d", &a[i]);
          //前缀和
          x1[1] = y1[1] = z1[1] = a[1];
          for(int i = 2; i <= n; i++){
               x1[i] = a[i] & x1[i-1];
               y1[i] = a[i] | y1[i-1];
               z1[i] = a[i] ^ z1[i-1];
          }
          //后缀和
          x2[n] = y2[n] = z2[n] = a[n];
          for(int i = n - 1; i >= 1; i--){
               x2[i] = a[i] & x2[i+1];
               y2[i] = a[i] | y2[i+1];
               z2[i] = a[i] ^ z2[i+1];
          }
          for(int i = 1; i <= m; i++){
               int q;
               scanf("%d", &q);
               if(q == 1)
                    printf("%d %d %d\n", x2[q+1], y2[q+1], z2[q+1]);
               else if(q == n)
                    printf("%d %d %d\n", x1[q-1], y1[q-1], z1[q-1]);
               else{
                    printf("%d %d %d\n", x1[q-1] & x2[q+1],
                                         y1[q-1] | y2[q+1],
                                         z1[q-1] ^ z2[q+1]);
               }
          }
     }
     return 0;
}