2017ICPC/广西邀请赛1005（水）HDU6186

CS Course

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 593 Accepted Submission(s): 288


[align=left]Problem Description[/align]
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

.

[align=left]Input[/align]
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,⋯,pq

in q lines, 1≤pi≤n

for each i in range[1,q].

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

3 3

1 1 1

1
2
3

[align=left]Sample Output[/align]

1 1 0
1 1 0
1 1 0
题意 求n个整数（除去） 二进制位运算 和 或 异或的结果
解析 求出前后缀询问时直接前缀后缀计算
AC代码

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <queue>
#include <vector>
using namespace std;
const int maxn= 1e5 + 10;
const int inf = 0x3f3f3f3f;
typedef long long ll;
int n,m;
int sum1[maxn],sum2[maxn],sum3[maxn];
int rsum1[maxn],rsum2[maxn],rsum3[maxn];
int a[maxn];
int main(int argc, char const *argv[])
{
while(scanf("%d %d",&n,&m)==2)
{
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];
//cout<<rsum1[i]<<" "<<rsum2[i]<<" "<<rsum3[i]<<endl;
}
int q;
while(m--)
{
scanf("%d",&q);
if(q==1)
printf("%d %d %d\n",rsum1[2],rsum2[2],rsum3[2]);
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;
}