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
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 3Sample Output
1 1 0 1 1 0 1 1 0
第一次见这种题，对于位运算，我真的很迷。这次用前缀后缀写出来了。
还是经过学长说了一次才明白。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
typedef long long LL;
using namespace std;
long long a[100010];
long long p1[100010],s1[100010];// &的前缀，后缀
long long p2[100010],s2[100010];// | 的前缀，后缀
int main()
{
int n,q,p;
long long ans1;
while(scanf("%d %d",&n,&q)!=EOF){
ans1=0;
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]);
if(i==1){//算前缀
p1[i]=p2[i]=a[i];
}
else {
p1[i]=(p1[i-1]&a[i]);
p2[i]=(p2[i-1]|a[i]);
}
ans1=a[i]^ans1;
}
for(int i=n;i>=1;i--){//算后缀
if(i==n){
s1[i]=s2[i]=a[i];
}
else {
s1[i]=(s1[i+1]&a[i]);
s2[i]=(s2[i+1]|a[i]);
}
}
for(int i=1;i<=q;i++){
scanf("%d",&p);
if(p==1){
printf("%lld %lld %lld\n",s1[2],s2[2],ans1^a[p]);
}
else if(p==n){
printf("%lld %lld %lld\n",p1[p-1],p2[p-1],ans1^a[p]);
}
else {
printf("%lld %lld %lld\n",p1[p-1]&s1[p+1],p2[p-1]|s2[p+1],ans1^a[p]);
}
}
}
return 0;
}