HDU4902Nice boat (线段树区间更新+lazytag)

Problem Description

There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli.

Let us continue our story, z*p(actually you) defeat the 'MengMengDa' party's leader, and the 'MengMengDa' party dissolved. z*p becomes the most famous guy among the princess's knight party.

One day, the people in the party find that z*p has died. As what he has done in the past, people just say 'Oh, what a nice boat' and don't care about why he died.

Since then, many people died but no one knows why and everyone is fine about that. Meanwhile, the devil sends her knight to challenge you with Algorithm contest.

There is a hard data structure problem in the contest:

There are n numbers a_1,a_2,...,a_n on a line, everytime you can change every number in a segment [l,r] into a number x(type 1), or change every number a_i in a segment [l,r] which is bigger than x to gcd(a_i,x) (type 2).

You should output the final sequence.

 

 

Input

The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains a integers n.
The next line contains n integers a_1,a_2,...,a_n separated by a single space.
The next line contains an integer Q, denoting the number of the operations.
The next Q line contains 4 integers t,l,r,x. t denotes the operation type.

T<=2,n,Q<=100000
a_i,x >=0
a_i,x is in the range of int32(C++)

 

 

Output

For each test case, output a line with n integers separated by a single space representing the final sequence.
Please output a single more space after end of the sequence

 

 

Sample Input

   
   
   
   

    
    
    
    1
10
16807 282475249 1622650073 984943658 1144108930 470211272 101027544 1457850878 1458777923 2007237709 
10
1 3 6 74243042
2 4 8 16531729
1 3 4 1474833169
2 1 8 1131570933
2 7 9 1505795335
2 3 7 101929267
1 4 10 1624379149
2 2 8 2110010672
2 6 7 156091745
1 2 5 937186357
   
   
   
   

 

 

Sample Output

   
   
   
   

    
    
    
    16807 937186357 937186357 937186357 937186357 1 1 1624379149 1624379149 1624379149 
   
   
   
   

 

 

Author

WJMZBMR

 

 

Source

2014 Multi-University Training Contest 4

 

涉及到更新区间（成段更新）时，有一个技巧lazy_tag，在线段树子树加上tag，表示这个子树包含某些性质或者要进行某些操作，不用立即访问到这个子树的叶子，只有当下一次仍需访问这个子树下的节点时才继续进行。lazy_tag技巧可以用于下面几个方面：

1、区间加一个数。(这里的加可以变为任何满足区间加法的操作，比如区间异或，区间求GCD）。
2、区间变为一个数。
3、区间最值。
4、给出01序列，区间取反。
5、区间最长连续1的序列。
6、区间最长子段和。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

struct nod{
    int l,r;
    int s;
}a[400100];
void build(int root,int l,int r){
    a[root].l=l;
    a[root].r=r;
    a[root].s=-1;
    if(l==r){
        scanf("%d",&a[root].s);
        return ;
    }
    if(l<r){
        int mid=(l+r)>>1;
        build(root<<1,l,mid);
        build(root<<1|1,mid+1,r);
        if(a[root<<1].s==a[root<<1|1].s)
            a[root].s=a[root<<1].s;
    }
}

void update1(int root,int l,int r,int val){
    if(a[root].l==l&&a[root].r==r){
        a[root].s=val;
        return ;
    }
    if(a[root].s>=0){//lazy ЛјПл
        a[root<<1].s=a[root].s;
        a[root<<1|1].s=a[root].s;
        a[root].s=-1;
    }
    int mid=(a[root].l+a[root].r)>>1;
    if(r<=mid)
        update1(root<<1,l,r,val);
    else if(l>mid)
        update1(root<<1|1,l,r,val);
    else{
        update1(root<<1,l,mid,val);
        update1(root<<1|1,mid+1,r,val);
    }
}

int gcd(int a,int b){
    if(b)
        return gcd(b,a%b);
    return a;
}

void update2(int root,int l,int r,int   val){
    if(a[root].l==l&&a[root].r==r&&a[root].s>=0){
        if(a[root].s>val)
            a[root].s=gcd(a[root].s,val);
        return ;
    }
    if(a[root].s>=0){
        a[root<<1].s=a[root].s;
        a[root<<1|1].s=a[root].s;
        a[root].s=-1;
    }
    int mid=(a[root].l+a[root].r)>>1;
    if(r<=mid)
        update2(root<<1,l,r,val);
    else if(l>mid)
        update2(root<<1|1,l,r,val);
    else{
        update2(root<<1,l,mid,val);
        update2(root<<1|1,mid+1,r,val);
    }
}

void output(int root){
    if(a[root].s>=0){
        for(int i=a[root].l;i<=a[root].r;i++)
            printf("%d ",a[root].s);
        return ;
    }
    output(root<<1);
    output(root<<1|1);
}
int main()
{
    int q,t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
       // cout<<"1****"<<endl;
        build(1,1,n);
       // cout<<"2****"<<endl;
        scanf("%d",&q);
        int o,l,r;
        int x;
        while(q--){
            scanf("%d%d%d%d",&o,&l,&r,&x);
         //   cout<<"3****"<<endl;
            if(o==1){
                update1(1,l,r,x);
           //     cout<<"4***"<<endl;
            }
            else if(o==2)
                update2(1,l,r,x);
        }
        output(1);
        printf("\n");
    }
    return 0;
}