题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4902
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.
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++)
Please output a single more space after end of the sequence
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
16807 937186357 937186357 937186357 937186357 1 1 1624379149 1624379149 1624379149
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。
代码例如以下：
#include <cstdio>
#include <algorithm>
#include <iostream>
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，假设是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，由于gcd仅仅对该区间比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("
");
}
return 0;
}