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
4
题目大意：三种类型的操作。A操作单点更新，给某位置增加值v，R操作区间更新，整个区间都更新为值v，Q操作区间查询，查询区间内的素数个数。
解题思路：用线段树维护三个值，一个lazy标记，一个val表示点的值，一个pnum表示素数的个数。在单点查询的时候，用一个PushDown操作将lazy标记下放，同时将值下发。最后还要用PushUp操作在回溯的时候更新值。对于区间查询来说，在查询的时候将lazy下放。关键要写好PushDown和PushUp这两个操作。（今天比较坑，自己建树的时候，没有上推，QAQ）。
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int maxn=1e5+200;
const int maxv=1e7+200;
struct SegTree{
int val,pnum,lazy;
}segtrees[maxn*4];
int prime[maxv],pprime[maxv];
void getprime(){
int num_prime=0;
prime[0]=prime[1]=1;
for(int i=2;i<maxv;i++){
if(!prime[i])
pprime[num_prime++]=i;
for(int j=0;j<num_prime && i*pprime[j]<maxv;j++){
prime[i*pprime[j]]=1;//合数标为1，同时，prime[j]是合数i*prime[j]的最小素因子
if(!(i%pprime[j]))//即比一个合数大的质数和该合数的乘积可用一个更大的合数和比其小的质数相乘得到
break;
}
}
}
void PushUp(int rt){
segtrees[rt].pnum = segtrees[rt*2].pnum + segtrees[rt*2+1].pnum; //
}
void build(int rt,int L,int R){
segtrees[rt].val=0;
segtrees[rt].lazy=0;
segtrees[rt].pnum=0;
if(L==R){
int &v=segtrees[rt].val;
scanf("%d",&v);
if(prime[v]){
segtrees[rt].pnum=0;
}else{
segtrees[rt].pnum=1;
}
return ;
}
build(lson);
build(rson);
PushUp(rt);
} void PushDown(int rt,int L,int R){
if(segtrees[rt].lazy==1){ segtrees[rt].lazy=0;
segtrees[rt*2].lazy=1;
segtrees[rt*2+1].lazy=1; segtrees[rt*2].val=segtrees[rt].val;
segtrees[rt*2+1].val=segtrees[rt].val;
segtrees[rt].val=0;
}
}
void add(int rt,int L,int R,int pos,int _val){
if(L==R){
int &v=segtrees[rt].val;
v+=_val;
if(prime[v]){
segtrees[rt].pnum=0;
}else{
segtrees[rt].pnum=1;
}
return ;
}
PushDown(rt,L,R);
if(pos<=mid){
add(lson,pos,_val);
}else{
add(rson,pos,_val);
}
PushUp(rt);
}
void repla(int rt,int L,int R,int l_ran,int r_ran,int _val){
if(l_ran<=L&&R<=r_ran){
segtrees[rt].lazy=1;
segtrees[rt].val=_val;
int &v=segtrees[rt].val;
if(prime[v]){
segtrees[rt].pnum=0;
}else{
segtrees[rt].pnum=R-L+1;
}
return ;
}
PushDown(rt,L,R);
if(l_ran<=mid){
repla(lson,l_ran,r_ran,_val);
} if(r_ran>mid){
repla(rson,l_ran,r_ran,_val);
}
PushUp(rt);
}
int query(int rt,int L,int R,int l_ran,int r_ran){
if(l_ran<=L&&R<=r_ran){
return segtrees[rt].pnum;
}
PushDown(rt,L,R);
int ret=0;
if(l_ran<=mid){
ret+=query(lson,l_ran,r_ran);
}
if(r_ran>mid){
ret+=query(rson,l_ran,r_ran);
}
return ret;
}
int main(){
int T,n,m;
char opr[10];
getprime();
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
build(1,1,n);
int a,b,c;
for(int i=1;i<=m;i++){
scanf("%s",opr);
if(opr[0]=='A'){
scanf("%d %d",&a,&b);
add(1,1,n,b,a);
}else if(opr[0]=='R'){
scanf("%d %d %d",&a,&b,&c);
repla(1,1,n,b,c,a);
}else{
scanf("%d %d",&a,&b);
int ans=query(1,1,n,a,b);
printf("%d\n",ans);
}
}
}
return 0;
}