ZOJ 3911 Prime Query（线段树）

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

Sample Output

来自 <http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3911>

【题意】：

单点修改，区间修改，查询区间素数个数

【解题思路】：

维护线段树，结点内容为：Val-实时数值(只针对单点) Sum-区间素数个数 Lazy-区间修改后传递给子区间

关键点在于数值和素数标记的同时更新和传递，由于区间更新时没有直接更新到最底层，故单点查询也需要pushdown操作。这里选择将区间更新后的值直接下传，至于素数标记，下传后再进行判断并赋值。

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<cmath>

 #include<algorithm>

 #define mid(a,b) ((a+b)>>1)

 #define LL int

 #define maxn 110000

 #define IN freopen("in.txt","r",stdin);

 using namespace std;

 char is_prime[maxn*];

 void sieve()

 {

     int m=(int)sqrt((maxn*)+0.5);

     fill(is_prime,is_prime+(maxn*),);

     is_prime[]=is_prime[]=;

     for(int i=;i<=m;i++) if(is_prime[i])

         for(int j=i*i;j<(maxn*);j+=i) is_prime[j]=;

 }

 int n,q;

 LL num[maxn];

 struct Tree

 {

     int left,right;

     LL sum;

     LL val,lazy;

 }tree[maxn<<];

 /*递归建树*/

 void build(int i,int left,int right)

 {

     tree[i].left=left;

     tree[i].right=right;

     tree[i].lazy=;

     if(left==right){

         tree[i].val=num[left];

         tree[i].sum=(is_prime[num[left]]? :);

         return ;

     }

     int mid=mid(left,right);

     build(i<<,left,mid);

     build(i<<|,mid+,right);

     tree[i].sum=tree[i<<].sum+tree[i<<|].sum;

 }

 /*区间修改，标记下传：每当访问到当前结点的子节点时，下传标记*/

 void pushdown(int i)

 {

     if(tree[i].lazy){

         int tmp = (is_prime[tree[i].lazy]? :);

         tree[i<<].val=tree[i].lazy;

         tree[i<<|].val=tree[i].lazy;

         tree[i<<].lazy=tree[i].lazy;

         tree[i<<|].lazy=tree[i].lazy;

         tree[i<<].sum=(tree[i<<].right-tree[i<<].left+)*tmp;

         tree[i<<|].sum=(tree[i<<|].right-tree[i<<|].left+)*tmp;

         tree[i].lazy=; /*下传后清零*/

     }

 }

 /*区间修改，d为改变量*/

 void update(int i,int left,int right,LL d)

 {

     if(tree[i].left==left&&tree[i].right==right)

     {

         int tmp = (is_prime[d]? :);

         tree[i].sum=(right-left+)*tmp;

         tree[i].val=d;

         tree[i].lazy=d;

         return ;

     }

     pushdown(i);

     int mid=mid(tree[i].left,tree[i].right);

     if(right<=mid) update(i<<,left,right,d);

     else if(left>mid) update(i<<|,left,right,d);

     else

     {

         update(i<<,left,mid,d);

         update(i<<|,mid+,right,d);

     }

     tree[i].sum=tree[i<<].sum+tree[i<<|].sum;

 }

 /*单点修改，d为改变量*/

 void update(int i,int x,LL d)

 {

     if(tree[i].left==tree[i].right){

         tree[i].val+=d;

         tree[i].sum=(is_prime[tree[i].val]? :);

         return;

     }

     pushdown(i);

     int mid=mid(tree[i].left,tree[i].right);

     if(x<=mid) update(i<<,x,d);

     else update(i<<|,x,d);

     tree[i].sum=tree[i<<].sum+tree[i<<|].sum;

 }

 /*区间结果查询*/

 LL query(int i,int left,int right)

 {

     if(tree[i].left==left&&tree[i].right==right)

         return tree[i].sum;

     pushdown(i);

     int mid=mid(tree[i].left,tree[i].right);

     if(right<=mid) return query(i<<,left,right);

     else if(left>mid) return query(i<<|,left,right);

     else return query(i<<,left,mid)+query(i<<|,mid+,right);

 }

 int main(int argc, char const *argv[])

 {

     //IN;

     sieve();

     int t;scanf("%d",&t);

     while(t--)

     {

         scanf("%d %d",&n,&q);

         for(int i=;i<=n;i++) scanf("%d",&num[i]);

         build(,,n);

         while(q--)

         {

             char c;

             while(c=getchar()){

                 if(c=='A'||c=='R'||c=='Q') break;

             }

             if(c=='A'){

                 int v,l;

                 scanf("%d %d",&v,&l);

                 update(,l,v);

             }

             if(c=='R'){

                 int a,l,r;

                 scanf("%d %d %d",&a,&l,&r);

                 update(,l,r,a);

             }

             if(c=='Q'){

                 int l,r;

                 scanf("%d %d",&l,&r);

                 printf("%d\n", query(,l,r));

             }

         }

     }

     return ;

 }