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 sum .
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.
Examples 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
Note
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.
给一个序列
支持3种操作
1 u v 对于所有i u<=i<=v，输出a[i]的和
2 u v t 对于所有i u<=i<=v a[i]=a[i]%t
3 u v 表示a[u]=v（将v赋值给a[u]）
n,q<=1e5 a[i],t,v<=1e9
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
提示
对于a%=b，如果a>=b，那么a至少除以2，也就是一个数最多减小log次
题目大意：
输入n和m，表示有n个数和m个询问，对于每个询问有三种操作：
- u v 对于所有i u<=i<=v，输出a[i]的和
- u v t 对于所有i u<=i<=v a[i]=a[i]%t
- u v 表示a[u]=v（将v赋值给a[u]）
解题思路：
这道题是线段树的基础题，对于这三种操作，单点替换和区间求和直接套用模板就可以，对于区间内每一个点需要剪枝，不剪枝一定会TLE，我们需要建两个树，一个存放sum的值，另一个用于存放每个节点的max，取模时我们判断一下该节点的子节点最大值是不是小于要取模的数，如果小于则直接不用判断。AC代码：
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
using ll = long long;
const ll _max=1e5+50;
ll arr[_max],tree[4*_max],treem[4*_max];
int main()
{
void build_tree(int,int,int);
void update(int,int,int,int,int);
ll query(int,int,int,int,int);
void mod1(int,int,int,int,int,int);
int n,m;
scanf("%d%d",&n,&m);
for ( int i = 0; i < n; i++)
cin>>arr[i];
build_tree(0,0,n-1);
while(m--)
{
int k,a,b,c;
scanf("%d",&k);
if(k==1)
{
scanf("%d%d",&a,&b);
ll sum=query(0,0,n-1,a-1,b-1);
printf("%lld
",sum);
}
else if(k==2)
{
scanf("%d%d%d",&a,&b,&c);
mod1(0,0,n-1,a-1,b-1,c);
}
else
{
scanf("%d%d",&a,&b);
update(0,0,n-1,a-1,b);
}
}
//system("pause");
return 0;
}
void build_tree(int node ,int start ,int end)
{
if(start==end)
{
tree[node]=arr[start];
treem[node]=arr[start];
return;
}
int left_node = node*2+1;
int right_node = node*2+2;
int mid=(start+end)>>1;
build_tree(left_node,start,mid);
build_tree(right_node,mid+1,end);
tree[node]=tree[left_node]+tree[right_node];
treem[node]=max(treem[left_node],treem[right_node]);
}
void update(int node, int start, int end, int idx, int val)
{
if(start==end)
{
tree[node]=val;
treem[node]=val;
return;
}
int left_node = node*2+1;
int right_node = node*2+2;
int mid=(start+end)>>1;
if(idx>=start&&idx<=mid)
update(left_node,start,mid,idx,val);
else
update(right_node,mid+1,end,idx,val);
tree[node]=tree[left_node]+tree[right_node];
treem[node]=max(treem[left_node],treem[right_node]);
}
ll query(int node, int start, int end, int l, int r)
{
if(l<=start&&r>=end)
return tree[node];
int left_node=node*2+1;
int right_node=node*2+2;
int mid=(start+end)>>1;
ll suml,sumr;
suml=sumr=0;
if(l<=mid)
suml=query(left_node,start,mid,l,r);
if(r>mid)
sumr=query(right_node,mid+1,end,l,r);
return suml+sumr;
}
void mod1(int node, int start, int end, int l, int r,int mod)
{
if(treem[node]<mod)
return;//这里注意一下，用子节点的最大值去比较
else if(start==end)
{
tree[node]=tree[node]%mod;
treem[node]=treem[node]%mod;
return;
}
else
{
int left_node=node*2+1;
int right_node=node*2+2;
int mid=(start+end)>>1;
if(l<=mid)
mod1(left_node,start,mid,l,r,mod);
if(r>mid)
mod1(right_node,mid+1,end,l,r,mod);
tree[node]=tree[left_node]+tree[right_node];
treem[node]=max(treem[left_node],treem[right_node]);
}
}