ZOJ 3911 线段树 区间更新查找,求素数个数
Prime Query
Time Limit: 1 Second
Memory Limit: 196608 KB
You are given a simple task. Given a sequence A[i] with N numbers. You have to perform Q operations on the given sequence.
Here are the operations:
- A v l, add the value v to element with index l.(1<=V<=1000)
- R a l r, replace all the elements of sequence with index i(l<=i<=r) with a(1<=a<=10^6) .
- Q l r, print the number of elements with index i(l<=i<=r) and A[i] is a prime number
Note that no number in sequence ever will exceed 10^7.
Input
The first line is a signer integer T which is the number of test cases.
For each test case, The first line contains two numbers N and Q (1 <= N, Q <= 100000) - the number of elements in sequence and the number of queries.
The second line contains N numbers - the elements of the sequence.
In next Q lines, each line contains an operation to be performed on the sequence.
Output
For each test case and each query,print the answer in one line.
Sample Input
1 5 10 1 2 3 4 5 A 3 1 Q 1 3 R 5 2 4 A 1 1 Q 1 1 Q 1 2 Q 1 4 A 3 5 Q 5 5 Q 1 5
Sample Output
2 1 2 4 0 4Author: HUA, Yiwei
Source: ZOJ Monthly, October 2015
#include<bits/stdc++.h>
using namespace std;
const int Max=1e7;
const int Maxn=1e6;
int vis[Max+1000];
struct Tree
{
int num,lazy;
}Tr[2*Maxn];
int init()
{
memset(vis,0,sizeof(vis));
vis[0]=vis[1]=1;
int m=sqrt(Maxn);
for(int i=2;i<=m;i++)
{
if(!vis[i])
{
for(int j=i*i;j<=Maxn;j+=i)
vis[j]=1;
}
}
}
int PushUp(int l,int r,int ind) //向下更新(主要更新区间的统计值)
{
if(Tr[ind<<1].lazy==Tr[ind<<1|1].lazy)
{
Tr[ind].lazy=Tr[ind<<1].lazy;
}
else
{
Tr[ind].lazy=0;
}
Tr[ind].num=Tr[ind<<1|1].num+Tr[ind<<1].num;
}
int PushDown(int l,int r,int ind) //向下更新(主要更新lazy(代表延迟操作))
{
if(Tr[ind].lazy&&l!=r)
{
Tr[ind<<1].lazy=Tr[ind<<1|1].lazy=Tr[ind].lazy;
if(Tr[ind].num)
{
int mid=(l+r)>>1;
Tr[ind<<1].num=mid-l+1;
Tr[ind<<1|1].num=r-mid;
}
else
Tr[ind<<1|1].num=Tr[ind<<1].num=0;
Tr[ind].lazy=0;
}
}
int Build(int l,int r,int ind)
{
Tr[ind].lazy=0;
Tr[ind].num=0;
if(l==r)
{
scanf("%d",&Tr[ind].lazy);
Tr[ind].num=(!vis[Tr[ind].lazy]);
return 0;
}
int mid=(l+r)>>1;
Build(l,mid,ind<<1);
Build(mid+1,r,ind<<1|1);
PushUp(l,r,ind);
}
int Add(int l,int r,int ind,int d,int t)
{
PushDown(l,r,ind); //变换之前向下更新(点的更新)
if(l==t&&r==t)
{
Tr[ind].lazy+=d;
Tr[ind].num=(!vis[Tr[ind].lazy]);
return 0;
}
int mid=(l+r)>>1;
if(t<=mid)
Add(l,mid,ind<<1,d,t);
else
Add(mid+1,r,ind<<1|1,d,t);
PushUp(l,r,ind);
}
int UpData(int l,int r,int ind,int L,int R,int d)
{
if(L>r||R<l)
return 0;
if(L<=l&&R>=r)
{
Tr[ind].lazy=d;
Tr[ind].num=(!vis[d])*(r-l+1);
return 0;
}
PushDown(l,r,ind); //区间的更新,变换之后向下更新
int mid=(l+r)>>1;
if(L<=mid)
UpData(l,mid,ind<<1,L,R,d);
if(R>mid)
UpData(mid+1,r,ind<<1|1,L,R,d);
PushUp(l,r,ind); //最后进行向上更新
}
int Query(int l,int r,int ind,int L,int R)
{
if(l>R||r<L)
return 0;
PushDown(l,r,ind);
if(L<=l&&R>=r)
{
return Tr[ind].num;
}
int mid=(l+r)>>1;
int sum=0;
if(L<=mid)
{
sum+=Query(l,mid,ind<<1,L,R);
}
if(R>mid)
{
sum+=Query(mid+1,r,ind<<1|1,L,R);
}
PushUp(l,r,ind);
return sum;
}
int main()
{
int T,n,q,l,r,d,t;
char s[10];
init();
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&q);
Build(1,n,1);
while(q--)
{
scanf("%s",s);
if(s[0]=='A')
{
scanf("%d%d",&d,&t);//a[t]+d;
Add(1,n,1,d,t);
}
if(s[0]=='Q')
{
scanf("%d%d",&l,&r);
printf("%d\n",Query(1,n,1,l,r));
}
if(s[0]=='R')
{
scanf("%d%d%d",&d,&l,&r);
UpData(1,n,1,l,r,d);
}
}
}
}