CodeForces 339D 线段树 单点更新,记录深度
Description
Xenia the beginner programmer has a sequence a, consisting of 2n non-negative integers: a1, a2, ..., a2n. Xenia is currently studying bit operations. To better understand how they work, Xenia decided to calculate some value v for a.
Namely, it takes several iterations to calculate value v. At the first iteration, Xenia writes a new sequence a1 or a2, a3 or a4, ..., a2n - 1 or a2n, consisting of 2n - 1 elements. In other words, she writes down the bit-wise OR of adjacent elements of sequence a. At the second iteration, Xenia writes the bitwise exclusive OR of adjacent elements of the sequence obtained after the first iteration. At the third iteration Xenia writes the bitwise OR of the adjacent elements of the sequence obtained after the second iteration. And so on; the operations of bitwise exclusive OR and bitwise OR alternate. In the end, she obtains a sequence consisting of one element, and that element is v.
Let's consider an example. Suppose that sequence a = (1, 2, 3, 4). Then let's write down all the transformations (1, 2, 3, 4) → (1 or 2 = 3, 3 or 4 = 7) → (3 xor 7 = 4). The result is v = 4.
You are given Xenia's initial sequence. But to calculate value v for a given sequence would be too easy, so you are given additional m queries. Each query is a pair of integers p, b. Query p, b means that you need to perform the assignment ap = b. After each query, you need to print the new value v for the new sequence a.
Input
The first line contains two integers n and m(1 ≤ n ≤ 17, 1 ≤ m ≤ 105). The next line contains 2n integers a1, a2, ..., a2n(0 ≤ ai < 230). Each of the next m lines contains queries. The i-th line contains integers pi, bi(1 ≤ pi ≤ 2n, 0 ≤ bi < 230) — the i-th query.
Output
Print m integers — the i-th integer denotes value v for sequence a after the i-th query.
Sample Input
2 4 1 6 3 5 1 4 3 4 1 2 1 2
1 3 3 3
Hint
For more information on the bit operations, you can follow this link: http://en.wikipedia.org/wiki/Bitwise_operation
题意:执行编号 1|2 3|4 5|6 7|8....
让他们的结果两两之间执行 ^
然后循环两两之间执行| 然后^一直循环
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
#include<cmath>
#define lind ind<<1
#define rind ind<<1|1
const int Maxn=1000000;
using namespace std;
int Tr[4*Maxn];
int a[4*Maxn];
int Pushup(int ind,int ans)
{
if(ans==0)
return 0;
if(ans%2==1)
{
Tr[ind]=Tr[ind<<1]|Tr[ind<<1|1];
}
else
{
Tr[ind]=Tr[ind<<1]^Tr[ind<<1|1];
}
}
int Build(int l,int r,int ind,int ans)
{
if(l==r)
{
scanf("%d",&Tr[ind]);
//cout<<ans<<"#"<<ind<<endl;
return 0;
}
int mid=(l+r)>>1;
Build(l,mid,ind<<1,ans-1);
Build(mid+1,r,ind<<1|1,ans-1);
a[ind]=ans;
//cout<<ind<<"$"<<ans<<endl;
Pushup(ind,a[ind]);
}
int Query(int l,int r,int ind,int t,int d)
{
if(l==r)
{
Tr[ind]=d;
return 0;
}
int mid=(l+r)>>1;
if(t<=mid)
Query(l,mid,lind,t,d);
else
Query(mid+1,r,rind,t,d);
Pushup(ind,a[ind]);
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int mm;
mm=pow(2,n);
Build(1,mm,1,n);
//printf("%d\n",Tr[1]);
int t,d;
for(int i=0;i<m;i++)
{
scanf("%d%d",&t,&d);
Query(1,mm,1,t,d);
printf("%d\n",Tr[1]);
}
}
return 0;
}