D. Mishka and Interesting sum
time limit per test
memory limit per test
input
output
a1, a2, ..., an of n
m
Each query is processed in the following way:
- l and r (1 ≤ l ≤ r ≤ n) are specified — bounds of query segment.
- [l, r] (in sequence of integers al, al + 1, ..., ar) even number of times, are written down.
- x1, x2, ..., xk, then Mishka wants to know the value , where — operator of exclusive bitwise OR.
Since only the little bears know the definition of array beauty, all you are to do is to answer each of queries presented.
Input
n (1 ≤ n ≤ 1 000 000) — the number of elements in the array.
n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — array elements.
m (1 ≤ m ≤ 1 000 000) — the number of queries.
m lines describes corresponding query by a pair of integers l and r (1 ≤ l ≤ r ≤ n) — the bounds of query segment.
Output
m
Examples
input
3 3 7 8 1 1 3
output
0
input
7 1 2 1 3 3 2 3 5 4 7 4 5 1 3 1 7 1 5
output
0 3 1 3 2
Note
In the second sample:
0.
3 is presented even number of times — the answer is 3.
1 is written down — the answer is 1.
1 and 2 are presented there even number of times. The answer is
.1 and 3 are written down. The answer is
. 【题意】
输入一个序列,有m次询问,询问任意区间段的出现次数为偶数的数字的异或之和。
【分析】
易知,某段区间的内数全部异或起来,得到的出现奇数次的数字异或和,那我们把每个出现过的数字都少(或多)异或一次,就得到了出现偶数次的数字异或和。
例如区间[1,R],预处理前缀异或和 s[i]表示前i个数字的符合题意的异或和,所以,当数字是第一次出现时,让他不参与异或,就达到目的了。这样S[i]就是[1,R]区间内的偶数次数字异或和。
但是上述做法只能做左区间是1的情况,推广到任意L,需要从1往后推,L++的同时,把当前数字x异或进前缀和(x的下一次出现的位置)相当于删掉当前数字后,下一次出现的位置成了第一次出现的位置。 树状数组维护S【i】非常合适。
把询问按左端点排序,依次查询的同时,删掉左边已经越过的部分。
【代码】
#include<bits/stdc++.h>
using namespace std;
const int N=1010101;
struct node{
int id,l,r,ans;
}e[N]; //询问
bool cmp1(node a,node b){
return a.l<b.l;
}
bool cmp2(node a,node b){
return a.id<b.id;
}
int n,m,a[N],Next[N];//next[i]:i处数字的下一个相同数字出现的地方
int c[N]; //BIT维护以L开头的偶数次数^和
void add(int k,int val)
{
for(;k<=n;k+=k&-k)
c[k]^=val;
}
int read(int k)
{
int res=0;
for(;k;k-=k&-k)res^=c[k];
return res;
}
int main()
{
while(cin>>n)
{
map<int,int>M; //出现次数
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(M[a[i]]++) //所有数都不算第一次
add(i,a[i]);
}
M.clear();
for(int i=n;i;i--) //为Next打表
{
Next[i]=M[a[i]];
if(Next[i]==0)Next[i]=n+1;
M[a[i]]=i;
}
cin>>m;
for(int i=0;i<m;i++)
scanf("%d%d",&e[i].l,&e[i].r),e[i].id=i;
sort(e,e+m,cmp1);
for(int i=0,l=1;i<m;i++)
{
while(l<e[i].l){add(Next[l],a[l]);l++;}//BIT右移
e[i].ans=read(e[i].r);
}
sort(e,e+m,cmp2);
for(int i=0;i<m;i++)
printf("%d\n",e[i].ans);
}
}