zoj3911 线段树区间修改与求素数的结合

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

题意：给定n个数，两种操作，修改和添加。每次询问l到r之间的素数的个数。
分析：由于输入的数据有空格，所以慎用getchar（）。。用scanf("%s")，就行啦。线段树还不够熟悉，足足写了一个多小时。。

#include<bitset>
#include<map>
#include<vector>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#define inf 0x3f3f3f3f
#define mem(a,x) memset(a,x,sizeof(a))

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

inline int in()
{
int res=0;char c;
while((c=getchar())<'0' || c>'9');
while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
return res;
}
const int N=10000100;

bitset<N> vis;

struct st
{
int l,r,flag,num,val;
}a[400040];

inline void PushUp(int i)
{
a[i].num=a[i<<1].num+a[i<<1|1].num;
}

inline void PushDown(int i)
{
if(a[i].flag)
{
a[i<<1].flag=a[i<<1|1].flag=1;
a[i<<1].val=a[i<<1|1].val=a[i].val;
int t=a[i].val;
bool f;
if(t%2==0 && t!=2)f=0;
else if(!vis[t]) f=1;
else f=0;
if(f)
{
a[i<<1].num=a[i<<1].r-a[i<<1].l+1;
a[i<<1|1].num=a[i<<1|1].r-a[i<<1|1].l+1;
}
else
{
a[i<<1].num=a[i<<1|1].num=0;
}
a[i].flag=0;
}
}
void build(int l,int r,int i)
{
a[i].l=l;
a[i].r=r;
a[i].flag=0;
if(l==r)
{
a[i].val=in();
if(a[i].val%2==0 && a[i].val!=2) a[i].num=0;
else{
if(!vis[a[i].val]) a[i].num=1;
else a[i].num=0;
}
return;
}
int mid=(l+r)>>1;
build(l,mid,i<<1);
build(mid+1,r,i<<1|1);
PushUp(i);
}
void update(int l,int r,int val,int i,char c)
{
if(a[i].l>=l && a[i].r<=r)
{
a[i].flag=1;

if(c=='a') a[i].val += val;
else a[i].val=val;

bool f;
if(a[i].val%2==0 && a[i].val!=2) f=0;
else if(!vis[a[i].val]) f=1;
else f=0;
if(f) a[i].num=a[i].r-a[i].l+1;
else a[i].num=0;
return;
}
PushDown(i);
int mid=(a[i].l+a[i].r)>>1;
if(l>mid)
{
update(l,r,val,i<<1|1,c);
}
else if(r<=mid)
{
update(l,r,val,i<<1,c);
}
else
{
update(l,mid,val,i<<1,c);
update(mid+1,r,val,i<<1|1,c);
}
PushUp(i);
}
int query(int l,int r,int i)
{
if(a[i].l>=l && a[i].r<=r) return a[i].num;
if(a[i].l>r || a[i].r<l) return 0;
PushDown(i);
return query(l,r,i<<1)+query(l,r,i<<1|1);
}
int main()
{
vis[0]=vis[1]=1;
for(int i=3;i*i<N;i+=2)
{
if(!vis[i]) for(int j=i*i;j<N;j+=i<<1) vis[j]=1;
}

int T=in();
while(T--)
{
int n=in(),q=in();
build(1,n,1);
char cmd[3];
while(q--)
{
scanf("%s",cmd);
if(cmd[0]=='A')
{
int val=in(),pos=in();
update(pos,pos,val,1,'a');
}
else if(cmd[0]=='R')
{
int val=in(),l=in(),r=in();
update(l,r,val,1,'r');
}
else
{
int l=in(),r=in();
printf("%d\n",query(l,r,1));

}
}
}
return 0;
}