Codeforces Round #250 (Div. 1) D. The Child and Sequence 线段树区间求和+点修改+区间取模

提示：扫一扫查出行
【扫一扫了解最新限行尾号】
复制提示

D. The Child and Sequence

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 test(s)

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题意：
操作1：l,r 求l到r之间的区间和
操作2：l,r,c 将l到r之间的数分别取模x
操作3：l,r 讲a[l]改成r；
题解：
线段树的区间操作
对于去摸：我们设置一个区间段最大值，要取得模要是大于这个最大则之间退出
这就是优化

///
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define inf 1000000007
#define mod 1000000007
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){
if(ch=='-')f=-;ch=getchar();
}
while(ch>=''&&ch<=''){
x=x*+ch-'';ch=getchar();
}return x*f;
}
//************************************************
const int maxn=+;
int a[maxn],n,m,q;
struct ss{
int l,r;
ll v,sum,tag;
}tr[maxn*];
void build(int k,int s,int t)
{
tr[k].l=s;tr[k].r=t;
tr[k].v=;tr[k].tag=;tr[k].sum=;
if(s==t){
tr[k].sum=a[s];
tr[k].v=a[s];
return ;
}
int mid=(s+t)>>;
build(k<<,s,mid);
build(k<<|,mid+,t);
tr[k].sum=tr[k<<].sum+tr[k<<|].sum;
tr[k].v=max(tr[k<<].v,tr[k<<|].v);
}
ll ask(int k,int s,int t)
{
if(tr[k].l==s&&tr[k].r==t){
return tr[k].sum;
}
int mid=(tr[k].l+tr[k].r)>>;
ll ret=;
if(t<=mid)ret= ask(k<<,s,t);
else if(s>mid)ret= ask(k<<|,s,t);
else {
ret=(ask(k<<,s,mid)+ask(k<<|,mid+,t));
} tr[k].sum=tr[k<<].sum+tr[k<<|].sum;
tr[k].v=max(tr[k<<].v,tr[k<<|].v);
return ret;
}
void modify(int k,int s,int t,ll c)
{
if(tr[k].v<c)return ;
if(tr[k].l==tr[k].r){
tr[k].sum%=c;
tr[k].v%=c;
return ;
}
int mid=(tr[k].l+tr[k].r)>>;
if(t<=mid)modify(k<<,s,t,c);
else if(s>mid)modify(k<<|,s,t,c);
else {
modify(k<<,s,mid,c);modify(k<<|,mid+,t,c);
}
tr[k].sum=tr[k<<].sum+tr[k<<|].sum;
tr[k].v=max(tr[k<<].v,tr[k<<|].v);
}
void update(int k,int x,int c)
{
if(tr[k].l==x&&tr[k].r==x){
tr[k].sum=c;
tr[k].v=c;
return;
}
int mid=(tr[k].l+tr[k].r)>>;
if(x<=mid)update(k<<,x,c);
else {
update(k<<|,x,c);
}
tr[k].sum=tr[k<<].sum+tr[k<<|].sum;
tr[k].v=max(tr[k<<].v,tr[k<<|].v);
}
int main(){ n=read();m=read();
for(int i=;i<=n;i++){
scanf("%d",&a[i]);
}build(,,n);int x,y;ll c;
for(int i=;i<=m;i++){
scanf("%d",&q);
if(q==){
scanf("%d%d",&x,&y);
printf("%I64d\n",ask(,x,y));
}
else if(q==){
scanf("%d%d%I64d",&x,&y,&c);
modify(,x,y,c);
}
else {
scanf("%d%d",&x,&y);
update(,x,y);
} }
return ;
}