ZOJ3911 Prime Query线段树

最新推荐文章于 2020-06-26 16:36:31 发布

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最新推荐文章于 2020-06-26 16:36:31 发布

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分类专栏： acm 文章标签：线段树

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本文详细介绍了一道涉及线段树区间修改与单点更新的经典算法题，通过具体实例讲解了如何利用线段树高效处理区间加法、区间替换及查询区间内素数数量的操作，适合初学者理解线段树的基本应用。

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链接
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
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
线段树入门题涉及到区间修改单点更新

#include <stdio.h>
#include <string.h>
const int maxn = 1e5+10;
const int maxsize = 1e7+1024;
struct ST
{
    int l,r;
    int sum;
    int add;//值
    int lazy;
} t[maxn*4];
int prime[maxsize];
int a;
void get()//打一个素数表
{
    memset(prime,0,sizeof(prime));
    prime[0] = prime[1] = 1;
    for(int i = 2; i < 3200; i++)
    {
        if(!prime[i])
        {
            for(int j = i*i; j < maxsize; j += i)
            {
                prime[j] = 1;   //0为素数
            }
        }
    }
}
void push_up(int p)
{
    t[p].sum = t[p<<1].sum+t[p<<1|1].sum;
}
/*向下打标记*/
void push_down(int p)
{
    if(t[p].lazy)
    {
        int tmp = (!prime[t[p].add])?1:0;
        t[p<<1].sum = tmp*(t[p<<1].r-t[p<<1].l+1);
        t[p<<1|1].sum = tmp*(t[p<<1|1].r-t[p<<1|1].l+1);
        t[p<<1|1].lazy = t[p<<1].lazy = t[p].lazy;
        t[p<<1].add = t[p].add;
        t[p<<1|1].add = t[p].add;
        t[p].lazy = 0;//最开始这里写成了t[p].add = 0; 而 t[p].add 会影响后面的值
    }
}
void build(int p,int l,int r)
{
    t[p].l = l;
    t[p].r = r;
    t[p].lazy = 0;
    if(l == r)
    {
        scanf("%d",&a);
        t[p].add = a;
        t[p].sum = (!prime[a])?1:0;
        return ;
    }
    int mid = (l+r)>>1;
    build(p<<1,l,mid);
    build(p<<1|1,mid+1,r);
    push_up(p);
}
void updata1(int p,int l,int r,int v)//区间更新
{
    if(l == t[p].l && r == t[p].r)
    {
        t[p].lazy = t[p].add = v;
        int tmp = (!prime[v])?1:0;
        t[p].sum = tmp*(t[p].r-t[p].l+1);
        return ;
    }
    push_down(p);
    int mid = (t[p].l+t[p].r)>>1;
    if(l > mid)
      updata1(p<<1|1,l,r,v);
    else if(r <= mid)
        updata1(p<<1,l,r,v);
    else
       {updata1(p<<1|1,mid+1,r,v);updata1(p<<1,l,mid,v);}
    push_up(p);
}
void updata(int p,int x,int v)//单点修改
{
    if(t[p].l == t[p].r)
    {
        t[p].add += v;
        t[p].sum = (!prime[t[p].add])?1:0;
        return ;
    }
    push_down(p);
    int mid = (t[p].l + t[p].r) >> 1;
    if(x <= mid)
        updata(p<<1,x,v);
    else
        updata(p<<1|1,x,v);
    push_up(p);
}
int ask(int p,int l,int r)
{
    if(l <= t[p].l && r >= t[p].r)
        return t[p].sum;
    push_down(p);
    int mid = (t[p].l+t[p].r)>>1;
    int val = 0;
    if(l <= mid)
        val += ask(p<<1,l,r);
    if(r > mid)
        val += ask(p<<1|1,l,r);
    /*if(l > mid)
        val += ask(p<<1|1,l,r);
    else if(r <= mid)
        val += ask(p<<1,l,r);
    else
        val = val + ask(p<<1|1,mid+1,r)+ask(p<<1,l,mid);*/
    return val;
}
void debug(int p,int l,int r)
{
    if(l == r)
    {
        printf("%d %d\n",t[p].add,t[p].sum);
        return ;
    }
    int mid = (l+r)>>1;
    if(l <= mid)
        debug(p<<1,l,mid);
    if(r > mid)
        debug(p<<1|1,mid+1,r);
}
int main()
{
    get();
    int t;
    scanf("%d",&t);
    char str[2];
    int v,l,r;
    while(t--)
    {
        int n,k;
        scanf("%d%d",&n,&k);
        build(1,1,n);
      //  debug(1,1,n);
   //  puts("");
        while(k--)
        {
            scanf("%s",str);
            if(str[0] == 'A')
            {
                scanf("%d%d",&v,&l);
                updata(1,l,v);
            }
            else if(str[0] == 'Q')
            {
                scanf("%d%d",&l,&r);
                printf("%d\n",ask(1,l,r));
            }
            else
            {
                scanf("%d%d%d",&v,&l,&r);
                updata1(1,l,r,v);
            }
           //    debug(1,1,n);
           // puts("");
        }
    }
    return 0;
}