hdu 4902 Nice boat（线段树区间修改，输出最终序列）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4902

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 

题目意思：
给出长度为n的数组，m个操作，操作有两种：1、1, l, r, x即把[l,r]段的元素全变为x。2、2,l,r,x即把[l,r]段的大于x的元素全变成该元素与x的最大公约数。

代码如下：

#include 
#include 
#include 
using namespace std;
#define LL long long
const int maxn = 111111;
#define ls (rt << 1)
#define rs (rt << 1 | 1)
#define lson l, mid, ls
#define rson mid+1, r, rs
//num数组表示该区间是否都是同一个数，如果是num就等于这个数，否则等-1
int num[maxn << 2];
//flag标记，表示该区间是否应该被纯色化（变成一个数）。
int flag[maxn << 2];
//mmax数组表示该区间的最大值，因为gcd只对该区间比x大的起作用
int mmax[maxn << 2];
int GCD(int a, int b)
{
	return a ? GCD(b%a,a) : b;
}
void Pushup(int rt)
{
	if(num[ls] == num[rs])
		num[rt] = num[ls];
	else
		num[rt] = -1;
	mmax[rt] = max(mmax[ls], mmax[rs]);
}

void Pushdown(int rt)
{
	if(flag[rt] != -1)
	{ 
		flag[ls] = flag[rs] = flag[rt];
		mmax[ls] = mmax[rs] = mmax[rt];
		num[ls] = num[rs] = num[rt];
		flag[rt] = -1;
	}
}

void build(int l, int r, int rt)
{
	flag[rt] = -1;
	if(l == r)
	{
		scanf("%d",&num[rt]);
		mmax[rt] = num[rt];//初始化
		return ;
	}
	int mid = (l+r) >> 1;
	build(lson);
	build(rson);
	Pushup(rt);
}

void update(int L, int R, int x, int l, int r, int rt)
{
	if(L <= l && r <= R)
	{
		flag[rt] = num[rt] = mmax[rt] = x;
		return ;
	}
	Pushdown(rt);
	int mid = (l+r) >> 1;
	if(L <= mid)
		update(L, R, x, lson);
	if(mid < R)
		update(L, R, x, rson);
	Pushup(rt);
}

void modify(int L, int R, int x, int l, int r, int rt)
{
	if(L <= l && r <= R && num[rt] > x)
	{
		flag[rt] = num[rt] = mmax[rt] = GCD(num[rt], x);
		return ;
	}
	Pushdown(rt);
	int mid = (l+r) >> 1;
	if(L <= mid && mmax[ls] > x)
		modify(L, R, x, lson);
	if(mid < R && mmax[rs] > x)
		modify(L, R, x, rson);
	Pushup(rt);
}

void Cout(int l, int r, int rt)
{
	if(l == r)
	{
		printf("%d ",num[rt]);
		return ;
	}
	Pushdown(rt);
	int mid = (l+r) >> 1;
	Cout(lson);
	Cout(rson);
}

int main()
{ 
	int T;
	int R, L;
	int n, m;
	int op,l, r, x;
	int i, j, k;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		build(1,n,1);
		scanf("%d",&m);
		for(i = 0; i < m; i++)
		{
			scanf("%d%d%d%d",&op,&l,&r,&x);
			if(op == 1)
			{
				update(l, r, x, 1, n, 1);
			}
			else if(op == 2)
			{
				modify(l, r, x, 1, n, 1);
			}
		}
		Cout(1, n, 1);
		printf("\n");
	}
	return 0;
}