The Child and Sequence（线段树）

The Child and Sequence

Time Limit: 4000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

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Description

At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations.
An operation can be one of the following:

Print operation l, r. Picks should write down the value of

.
Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for
each i(l ≤ i ≤ r).
Set operation k, x. Picks should set the value of a[k] to x (in
other words perform an assignment a[k] = x).

Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m(1 ≤ n, m ≤ 105). The second line
contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) —
initial value of array elements.

Each of the next m lines begins with a number type

.

If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond
the operation 1.
If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109),
which correspond the operation 2.
If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109),
which correspond the operation 3.

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Sample Input

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3

Output

8
5

Input

10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10

Output

49
15
23
1
9

Hint

Consider the first testcase:

At first, a = {1, 2, 3, 4, 5}.
After operation 1, a = {1, 2, 3, 0, 1}.
After operation 2, a = {1, 2, 5, 0, 1}.
At operation 3, 2 + 5 + 0 + 1 = 8.
After operation 4, a = {1, 2, 2, 0, 1}.
At operation 5, 1 + 2 + 2 = 5.

代码：

#include<iostream>
using namespace std;
#define MAXN 100000

long long tree[MAXN*4+9];
int mv[MAXN*4+9];

void build(int s,int t,int start)
{
if(s==t)
{
cin>>tree[start];
mv[start]=tree[start];
return ;
}
int mid=(s+t)/2;
build(s,mid,start*2);
build(mid+1,t,start*2+1);
tree[start]=tree[start*2]+tree[start*2+1];
mv[start]=max(mv[start*2],mv[start*2+1]);
}

void sets(int k,int x,int s,int t,int start)
{
if(s==t)
{
tree[start]=x;
mv[start]=x;
return ;
}
int mid=(s+t)/2;
if(k<=mid) sets(k,x,s,mid,start*2);
else sets(k,x,mid+1,t,start*2+1);
tree[start]=tree[start*2]+tree[start*2+1];
mv[start]=max(mv[start*2],mv[start*2+1]);
}

void update(int ll,int rr,int x,int s,int t,int start)
{
if(s==t)
{
tree[start]=tree[start]%x;
mv[start]=tree[start];
return ;
}
if(mv[start]<x) return ;
int mid=(s+t)/2;
if(ll<=mid) update(ll,rr,x,s,mid,start*2);
if(rr>mid) update(ll,rr,x,mid+1,t,start*2+1);
tree[start]=tree[start*2]+tree[start*2+1];
mv[start]=max(mv[start*2],mv[start*2+1]);
}

long long query(int ll,int rr,int s,int t,int start)
{
if(ll<=s&&rr>=t)
return tree[start];
long long sum=0;
int mid=(s+t)/2;
if(ll<=mid) sum+=query(ll,rr,s,mid,start*2);
if(rr>mid) sum+=query(ll,rr,mid+1,t,start*2+1);
return sum;
}

int main()
{
int n,m,type,ll,rr,k,x;
while(cin>>n>>m)
{
build(1,n,1);
for(int i=1; i<=m; i++)
{
cin>>type;
if(type==1)
{
cin>>ll>>rr;
cout<<query(ll,rr,1,n,1)<<endl;
}
if(type==2)
{
cin>>ll>>rr>>x;
update(ll,rr,x,1,n,1);
}
if(type==3)
{
cin>>k>>x;
sets(k,x,1,n,1);
}
}
}
return 0;
}