HDU

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 

题意：1操作代表改区间为x，2操作代表对于区间假设这个数>x的话就改成两个数的gcd

思路：线段树区间改动。队友给的思路。多一个flag代表的是这个区间的数是否都是一样的

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#define lson(x) ((x) << 1)
#define rson(x) ((x) << 1 | 1)
typedef long long ll;
using namespace std;
const int MAXN = 100005;
const int ROOT = 1;

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

struct seg {
	int w;
	int v;
	int flag;
};

struct segment_tree {
	seg node[MAXN << 2];

	void update(int pos) {
		node[pos].flag = (node[lson(pos)].flag && node[rson(pos)].flag && node[lson(pos)].w == node[rson(pos)].w);
		node[pos].w = node[lson(pos)].w;
	}

	void build(int l, int r, int pos) {
		node[pos].flag = 0;
		node[pos].v = -1;
		if (l == r) {
			scanf("%d", &node[pos].w);
			node[pos].flag = 1;
			return;
		}
		int m = l + r >> 1;
		build(l, m, lson(pos));
		build(m + 1, r, rson(pos));
		update(pos);
	}

	void push(int l, int r, int pos) {
		if (node[pos].v != -1) {
			node[lson(pos)].v = node[rson(pos)].v = node[pos].v;
			node[lson(pos)].w = node[rson(pos)].w = node[pos].v;
			node[pos].v = -1;;
		}
	}

	void modify(int l, int r, int pos, int x, int y, int z) {
		if (x <= l && r <= y) {
			node[pos].w = z;
			node[pos].v = z;;
			/* node[pos].flag = z; is wrong. */
			return;
		}
		push(l, r, pos);
		int m = l + r >> 1;
		if (x <= m) modify(l, m, lson(pos), x, y, z);
		if (y > m) modify(m + 1, r, rson(pos), x, y, z);
		update(pos);
	}

	void modify1(int l, int r, int pos, int x, int y, int z) {
		if (node[pos].flag && node[pos].w <= z)
			return;
		if (x <= l && y >= r && node[pos].flag) {
			node[pos].w = gcd(node[pos].w, z);
			node[pos].v = node[pos].w;
			return;
		}
		push(l, r, pos);
		int m = l + r >> 1;
		if (x <= m) modify1(l, m, lson(pos), x, y, z);
		if (y > m) modify1(m + 1, r, rson(pos), x, y, z);
		update(pos);
	}

	int query(int l, int r, int pos, int x, int y) {
		if (x <= l && r <= y) return node[pos].w;
		push(l, r, pos);
		int m = l + r >> 1;
		int res = 0;
		if (x <= m) res += query(l, m, lson(pos), x, y);
		if (y > m) res += query(m + 1, r, rson(pos), x, y);
		return res;
	}
} tree;

int main() {
	int t, n, m;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		tree.build(1, n, 1);
		scanf("%d", &m);
		int op, l, r, x;
		while (m--) {
			scanf("%d%d%d%d", &op, &l, &r, &x);
			if (op == 1)
				tree.modify(1, n, 1, l, r, x);
			else tree.modify1(1, n, 1, l, r, x);
		}
		for (int i = 1; i <= n; i++)
			printf("%d ", tree.query(1, n, 1, i, i));
		printf("
");
	}
	return 0;
}