ZOJ 3911 Prime Query (线段树基本操作)

Prime Query

Time Limit: 5 Seconds 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

Sample Output

Author: HUA, Yiwei
Source: ZOJ Monthly, October 2015

#include<bits/stdc++.h>
using namespace std;

#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)
#define read(x,y) scanf("%d%d",&x,&y)

#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define root l,r,rt
#define lrt int l,int r,int rt
/*
题目大意:线段树基本套路，
区间整体修改，单点更新，没什么特别的。

先预处理素数，
这里本人踩了个坑，，，
筛法时j从i*i开始是有条件的。。。注意数据范围。
懒惰标记设定为要修改成的那个值，其余的都是基础训练了。


*/
const int  maxn =1e6+5;
///素数筛法
int vis[maxn];
///数据域部分
int n,m,t;
int x,y,z;
int a[maxn];
char  s[5];
///线段树
int lazy[maxn<<2],sum[maxn<<2],tree[maxn<<2];///tree代表数据区间和，sum代表答案区间和
void pushup(lrt)
{
    tree[rt]=tree[rt<<1]+tree[rt<<1|1];
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(lrt)
{
    if(l==r) return;
    int mid=l+r>>1;
    if(lazy[rt])///懒惰标记记录了区间修改值
    {
        lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt];

        tree[rt<<1]=(mid-l+1)*lazy[rt];
        tree[rt<<1|1]=(r-mid)*lazy[rt];

        sum[rt<<1]=(mid-l+1)*vis[lazy[rt]];
        sum[rt<<1|1]=(r-mid)*vis[lazy[rt]];
        lazy[rt]=0;
    }
}
void build(lrt)
{
    lazy[rt]=0;
    if(l==r)
    {
        tree[rt]=a[l];
        sum[rt]=vis[a[l]];
        return ;
    }
    int mid=l+r>>1;
    build(lson),build(rson),pushup(root);
}
void update(lrt,int L,int R,int v)
{
    if(L<=l&&r<=R)
    {
        sum[rt]=(r-l+1)*vis[v];
        tree[rt]=(r-l+1)*v;
        lazy[rt]=v;
        return ;
    }
    pushdown(root);
    int mid=l+r>>1;
    if(L<=mid)  update(lson,L,R,v);
    if(mid<R) update(rson,L,R,v);
    pushup(root);
}
void update(lrt,int pos,int v)
{
    if(l==r)
    {
        tree[rt]+=v;
        sum[rt]=vis[tree[rt]];
        return ;
    }
    pushdown(root);
    int mid=l+r>>1;
    if(pos<=mid) update(lson,pos,v);
    if(mid<pos) update(rson,pos,v);
    pushup(root);
}
int query(lrt,int L,int R)
{
    if(L<=l&&r<=R)  return sum[rt];
    pushdown(root);
    int mid=l+r>>1,ans=0;
    if(L<=mid) ans+=query(lson,L,R);
    if(mid<R) ans+=query(rson,L,R);
    pushup(root);
    return ans;
}

int main()
{
    memset(vis,0,sizeof(vis));
    vis[1]=1;///1不是素数
    for(int i=2;i<maxn;i++)
    {
        ///cout<<i<<endl;
        if(vis[i]) continue;
        for(int j=i*2;j<maxn;j+=i)  vis[j]=1;///这里一个wa点，如果使用i*i可能会爆数据。。。
    }
    for(int i=1;i<maxn;i++) vis[i]^=1;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        build(1,n,1);
        for(int i=0;i<m;i++)
        {
            scanf("%s",s);
            if(s[0]=='A')
            {
                scanf("%d%d",&x,&y);
                update(1,n,1,y,x);
            }
            else if(s[0]=='Q')
            {
                scanf("%d%d",&x,&y);
                printf("%d\n",query(1,n,1,x,y));
            }
            else
            {
                scanf("%d%d%d",&z,&x,&y);
                update(1,n,1,x,y,z);
            }
        }
    }
    return 0;
}

ZOJ 3911 Prime Query (线段树基本操作)

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