fzu-2105 (segment tree update interval)

Meaning of the questions: interval has three binary operations, the value range of not more than 16, and then query interval. Since each number is small, so a large number of duplicates in the memory section, and after the operation interval equal to the binary value more and more. We only need to record the current value is equal to the interval.

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
int tree[1000000*4];
int a[1000000];
void push_down(int l,int r,int k)
{
	if(tree[k<<1]!=-1&&tree[k<<1|1]==tree[k<<1])
	{
		tree[k]=tree[k<<1];
	}
}
void build(int l,int r,int k)
{
	tree[k]=-1;
	if(l==r)
	{
		tree[k]=a[l];
		return ;
	}
	int mid=(l+r)>>1;
	build(l,mid,k<<1);
	build(mid+1,r,k<<1|1);
	push_down(l,r,k);
}
void update(int L,int R,int l,int r,int k,int flag,int val)
{
	int mid=(L+R)>>1;
	if(L>=l&&R<=r&&tree[k]!=-1)
	{
		if(flag==1)tree[k]^=val;
		else if(flag==2)tree[k]|=val;
		else if(flag==3)tree[k]&=val;
		return ;
	}
	if(tree[k]>=0)		//下传标记 
	{
		tree[k<<1]=tree[k<<1|1]=tree[k];tree[k]=-1;
	}
	if(l<=mid) update(L,mid,l,r,k<<1,flag,val);
	if(r>mid) update(mid+1,R,l,r,k<<1|1,flag,val);
	push_down(L,R,k);//上传标记 
}
int query(int L,int R,int l,int r,int k)
{
	int ans=0;
	int mid=(L+R)>>1;
	if(L>=l&&R<=r&&tree[k]!=-1)
	{
		return tree[k]*(R-L+1);
	}
	if(tree[k]>=0)		//下传标记 
	{
		tree[k<<1]=tree[k<<1|1]=tree[k];tree[k]=-1;
	}
	if(l<=mid)
	ans+=query(L,mid,l,r,k<<1);
	if(r>mid)
	ans+=query(mid+1,R,l,r,k<<1|1);
	push_down(L,R,k);//上传标记
	return ans;
}
int main()
{
	int t;scanf("%d",&t);
	while(t--)
	{
		int n,q;
		scanf("%d%d",&n,&q);
		for(int i=1;i<=n;i++) scanf("%d",&a[i]);
		build(1,n,1);
		while(q--)
		{
			char s[5];int x,y,z;
			scanf("%s",s);
			if(s[0]=='S')
			{
				scanf("%d%d",&x,&y);x++;y++;
				printf("%d\n",query(1,n,x,y,1));
			}
			else if(s[0]=='X')
			{
				scanf("%d%d%d",&z,&x,&y);x++;y++;
				update(1,n,x,y,1,1,z);		//orx
			}
			else if(s[0]=='O')
			{
				scanf("%d%d%d",&z,&x,&y);x++;y++;
				update(1,n,x,y,1,2,z);		//or
			}
			else
			{
				scanf("%d%d%d",&z,&x,&y);x++;y++;			//and
				update(1,n,x,y,1,3,z);
			}
		}
	}
	
	return 0;
}