Nice boat
Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1038 Accepted Submission(s):
465
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.
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++)
separated by a single space representing the final sequence.
Please output
a single more space after end of the sequence
最后输出变化后的一组数。
用线段树写的，主要是利用mark标记 ，全部变成x的话直接把这段变成x，然后加个tree[i].mark就行了，如果往下更新，要把标记和数一起往下传具体看pushdown函数，查询的时候如果延迟标记存在的话直接把这一段输出，因为标记存在这一段肯定是相同的数。。
第2步操作主要是，如果什么已经标记了则全都是相同的数，比较一下求公约数。
如果没标记就要tree[i].left==tree[i].right。比较一下这个数，就行了。
#include<iostream>
#include<cstring>
#include<cstdio>
const int maxx=;
using namespace std;
int a[*maxx];
struct tree
{
int right,left;
int mark,num;
}tree[*maxx];
int gcd(int a,int b)
{
return b==?a:gcd(b,a%b);
}
void build(int i,int left,int right)
{
tree[i].left=left;
tree[i].right=right;
tree[i].mark=;
if(left==right)
{
tree[i].num=a[left];
return ;
}
int mid=(left+right)/;
build(*i,left,mid);
build(*i+,mid+,right);
}
void pushdown(int r)
{
if(tree[r].mark)
{
tree[*r].num=tree[*r+].num=tree[r].num;
tree[*r].mark=tree[*r+].mark=tree[r].mark;
tree[r].mark=;
}
}
void update(int r,int i,int j,int k)
{
int mid;
mid=(tree[r].left+tree[r].right)/;
if(tree[r].left>=i&&tree[r].right<=j)
{
tree[r].num=k;
tree[r].mark=;
return;
}
pushdown(r);//向下传递。
if(i>mid)
update(*r+,i,j,k);
else if(j<=mid)
update(r*,i,j,k);
else
{
update(*r,i,mid,k);
update(*r+,mid+,j,k);
}
}
void change(int r,int i,int j,int x)
{
int mid;
mid=(tree[r].left+tree[r].right)/;
if(tree[r].mark)
{
if(i<=tree[r].left&&j>=tree[r].right)
{
if(tree[r].num<=x)
return ;
else
{
tree[r].num=gcd(tree[r].num,x);
return ;
}
}
}
if(tree[r].left==tree[r].right)
{
if(tree[r].num>x)
{
tree[r].num=gcd(tree[r].num,x);
return ;
}
else
return ;
}
pushdown(r);
if(i>mid)
change(*r+,i,j,x);
else if(j<=mid)
change(r*,i,j,x);
else
{
change(*r,i,mid,x);
change(*r+,mid+,j,x);
}
}
void prit(int r)
{
int i;
if(tree[r].left==tree[r].right)
{
printf("%d ",tree[r].num);
return ;
}//
if(tree[r].mark)
{
for(i=tree[r].left;i<=tree[r].right;i++)
printf("%d ",tree[r].num);
return;
} prit(*r);
prit(*r+); }
int main()
{
int t,n,x,y,m,k,v,i;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=;i<=n;i++)
scanf("%d",&a[i]); scanf("%d",&m);
build(,,n);
for(i=;i<=m;i++)
{
scanf("%d",&k);
if(k==)
{
scanf("%d%d%d",&x,&y,&v);
update(,x,y,v);
}
else
{
scanf("%d%d%d",&x,&y,&v);
change(,x,y,v);
} }
prit();
printf("\n");
}
return ;
}