HDU6186 | 2017广西邀请赛 CS Course （前缀和后缀）

CS Course

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 430    Accepted Submission(s): 222


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

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

思路：题目要求的删除第q个数候所有数的 & | ^和，所以提前求出前缀和后缀，每次& | ^ 前i-1个和后i+1个即可。

AC代码如下：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
int n,m,sum1[maxn],sum2[maxn],sum3[maxn],rsum1[maxn],rsum2[maxn],rsum3[maxn],a[maxn];

int main(){
while(~scanf("%d%d",&n,&m)){
scanf("%d",&a[1]);
sum1[1] = sum2[1] = sum3[1] = a[1];
for(int i = 2; i <= n; i ++){
scanf("%d",&a[i]);
sum1[i] = sum1[i-1] & a[i];
sum2[i] = sum2[i-1] | a[i];
sum3[i] = sum3[i-1] ^ a[i];
}
rsum1
= rsum2
= rsum3
= a
;
for(int i = n-1; i >= 1; i --){
rsum1[i] = rsum1[i+1] & a[i];
rsum2[i] = rsum2[i+1] | a[i];
rsum3[i] = rsum3[i+1] ^ a[i];
}
int q;
while(m --){
scanf("%d",&q);
if(q == 1) printf("%d %d %d\n",rsum1[q+1],rsum2[q+1],rsum3[q+1]);
else if(q == n) printf("%d %d %d\n",sum1[n-1],sum2[n-1],sum3[n-1]);
else printf("%d %d %d\n",sum1[q-1] & rsum1[q+1],sum2[q-1]|rsum2[q+1],sum3[q-1]^rsum3[q+1]);
}
}
return 0;
}