HDU6186 CS Course【位运算+前缀和+后缀和】

CS Course

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1964    Accepted Submission(s): 812

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 31 1 1123

 

Sample Output

1 1 01 1 01 1 0

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

问题链接：HDU6186 CS Course

问题简述：（略）

问题分析：

  这个问题出现了三种运算，分别是与、或和异或。

  查询时，输入k，求除了a[k]之外，所有a[i]与、或和异或的结果。这也是该问题要做的计算。

  异或运算比较好办，因为x^a^a = x，所以先将所有数异或起来，然后再计算。

  解法一：

  采用前缀和计算原理，先算出前缀与以及前缀或。同理，也算出后缀与以及后缀或。再计算结果就简单了。

  解法二：

  异或运算同解法一。

  与运算以及或运算，可以通过统计各个数的各位的1的个数之后，再行计算。根据与运算的性质，所有数的某位都为1时则那个位才为1，否则那位为0；同理，根据或运算的性质，所有数中，某位只要有1位为1则那个位为1，否则那位为0。

程序说明：（无）

题记：（略）

参考链接：（略）

AC的C++语言程序（解法一）如下：

/* HDU6186 CS Course */

#include <iostream>
#include <stdio.h>

using namespace std;

const int N = 1e5;
int a[N + 1], andl[N + 2], andr[N + 2], orl[N + 2], orr[N + 2], xora;

int main()
{
    int n, q;
    while(~scanf("%d%d", &n, &q)) {
        // 为计算前缀与，前缀或，后缀与，后缀或等进行初始化
        andl[0] = andr[n + 1] = 0;
        andl[0] = ~andl[0];
        andr[n + 1] = ~andr[n + 1];
        orl[0] = orr[n + 1] = 0;
        xora = 0;

        // 读入数据
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);

        // 计算前缀与、前缀或和异或值
        for(int i = 1; i <= n; i++) {
            andl[i] = andl[i - 1] & a[i];
            orl[i] = orl[i - 1] | a[i];
            xora ^= a[i];
        }
        // 计算后缀与，后缀或
        for(int i = n; i >= 1; i--) {
            andr[i] = andr[i + 1] & a[i];
            orr[i] = orr[i + 1] | a[i];
        }

        for(int i = 1; i <= q; i++) {
            int k;
            scanf("%d", &k);

            printf("%d %d %d\n", andl[k - 1] & andr[k + 1], orl[k - 1] | orr[k + 1], xora ^ a[k]);
        }
    }

    return 0;
}

AC的C++语言程序（解法二）如下：

/* HDU6186 CS Course */

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

const int N = 1e5;
const int BITS = 32;
int a[N + 1], xora, cnt1[BITS + 1], cnt[BITS + 1];

int main()
{
    int n, q;
    while(~scanf("%d%d", &n, &q)) {
        // 为计算异或和统计的初始化
        xora = 0;
        memset(cnt1, 0, sizeof(cnt1));

        // 读入数据, 计算异或值
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            xora ^= a[i];
        }
        // 统计各位1的个数
        for(int i = 1; i <= n; i++) {
            int t = a[i], j = 1;
            while(t) {
                if(t & 1)
                    cnt1[j]++;
                t >>= 1;
                j++;
            }
        }

        for(int i = 1; i <= q; i++) {
            int k;
            scanf("%d", &k);

            // 计算除了第k个数之外，各位的1的个数
            memcpy(cnt, cnt1, sizeof(cnt));
           int t = a[k], j = 1;
           while(t) {
               if(t & 1)
                   cnt[j]--;
               t >>= 1;
               j++;
           }
           // 计算与和或的结果
           int anda = 0, ora = 0;
           for(int i = BITS; i >= 1; i--) {
               anda <<= 1;
               if(cnt[i] == n - 1)
                   anda |= 1;
               ora <<= 1;
               if(cnt[i] > 0)
                   ora |= 1;
           }

            printf("%d %d %d\n", anda, ora, xora ^ a[k]);
        }
    }

    return 0;
}