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
這數據一眼看上去十分嚇人，但是仔細讀題之後發現就是 long long的線段樹
操作一：修改區間內所有的值爲 x
操作二：對給定區間的所有值詢問，如果大於等於x則改爲 gcd（a[i],x）,否則不修改
這道題的關鍵就是在lazy上, 操作一是將所有的值改成x,那麼我們的lazy可以存個x,放在祖先節點上,到時候pushdown的時候可以直接修改成祖先的值.
操作二:對於還沒有pushdown的祖先,我們也可以對此祖先直接求gcd,因爲反正改祖先的子節點都需要改的.
最後的pushup應該怎麼寫呢,爲了與pushdown對應,那麼我們就存下左右子節點相同的值,不相同存個-1.
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <stdio.h>
#include <ctype.h>
#include <bitset>
#define LL long long
#define ULL unsigned long long
#define mod 100000000
#define INF 0x7ffffff
#define mem(a,b) memset(a,b,sizeof(a))
#define MODD(a,b) (((a%b)+b)%b)
using namespace std;
const int maxn = 1e6 + 5;
int n,m;
LL a[maxn << 2];
LL T[maxn << 2];//這裏沒有寫lazy數組,但是lazy的思想在T[]的祖先節點上了
LL gcd(LL a,LL b){return b == 0 ? a : gcd(b,a%b);}
void pushUp(int rt)
{
if(T[rt << 1] == T[rt << 1 | 1]) T[rt] = T[rt << 1];//子節點相同,存下來
else T[rt] = -1;
}
void pushDown(int rt)
{
if(~T[rt]){//不爲-1 下推
T[rt << 1] = T[rt];
T[rt << 1 | 1] = T[rt];
T[rt] = -1;
}
return ;
}
void buildTree(int rt,int l,int r)
{
T[rt] = -1;
if(l == r){
T[rt] = a[l];
return;
}
int mid = (l + r) >> 1;
buildTree(rt << 1,l,mid);
buildTree(rt << 1 | 1,mid + 1,r);
pushUp(rt);
}
void upDatef(int rt,int l,int r,int L,int R,LL v)
{
if(L <= l && R >= r){
T[rt] = v;
return;
}
int mid = (l + r) >> 1;
pushDown(rt);
if(L <= mid) upDatef(rt << 1,l,mid,L,R,v);
if(R > mid) upDatef(rt << 1 | 1,mid + 1,r,L,R,v);
pushUp(rt);
}
void upDates(int rt,int l,int r,int L,int R,LL v)
{
if(L <= l && R >= r){
if(~T[rt]){
if(T[rt] >= v) T[rt] = gcd(T[rt],v);
return;
}
}
int mid = (l + r) >> 1;
pushDown(rt);
if(L <= mid) upDates(rt << 1,l,mid,L,R,v);
if(R > mid) upDates(rt << 1 | 1,mid + 1,r,L,R,v);
pushUp(rt);
}
void query(int rt,int l,int r)
{
if(l == r){
printf("%lld ",T[rt]);
return;
}
int mid = (l + r) >> 1;
pushDown(rt);
query(rt << 1,l,mid);
query(rt << 1 | 1,mid + 1,r);
}
int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i = 1; i <= n; i++) scanf("%lld",&a[i]);
buildTree(1,1,n);
int m;
scanf("%d",&m);
while(m--){
int x,y,opt;
LL k;
scanf("%d",&opt);
if(opt == 1){
int x,y;
LL v;
scanf("%d%d%lld",&x,&y,&v);
upDatef(1,1,n,x,y,v);
}
else if(opt == 2){
int x,y;
LL v;
scanf("%d%d%lld",&x,&y,&v);
upDates(1,1,n,x,y,v);
}
}
query(1,1,n);
printf("\n");
}
return 0;
}