Codeforces Round #250 (Div. 1) D. The child and Sequence segment tree interval summation + point modification + interval modulus

D. the Child and Sequence

At the children's, the child came to picks's house, and messed he 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 being one of the following:

1. Print Operation l, R. Picks should write down the value of.
2. Modulo Operation l, R, x. Picks should perform assignment a[i] = a[i] mod x For each i (l ≤ i ≤ r).
3. Set Operation k, x. Picks should set the value of a[K] to x (in and words perform an assignment c29>a[k] = x).

Can picks to perform the whole sequence of operations?

Input

The first line of input contains the 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'll be is the integers more in the line: l, R (1≤ l ≤ r< /c7>≤ n), which correspond the operation 1.
• If Type = 2, there'll is 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'll be is the 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 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


Test instructions
   Operation 1:l,r the interval between L and R and
   Action 2:l,r,c the  number from L to R is modulo x respectively
   Operation 3:l,r  talk a[l] change to R;
Exercises
   Interval operation of Segment tree
   For the go-to-touch: We set a maximum interval segment, to get the modulus if greater than the maximum between the exit
   This is the optimization

///1085422276#include <bits/stdc++.h>using namespacestd; typedefLong Longll;#defineMem (a) memset (A,0,sizeof (a))#defineMeminf (a) memset (A,127,sizeof (a));#defineINF 1000000007#defineMoD 1000000007inline ll read () {ll x=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){        if(ch=='-') f=-1; ch=GetChar (); }     while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar (); }returnx*F;}//************************************************Const intmaxn=100000+5;inta[maxn],n,m,q;structss{intL,r; ll V,sum,tag;} TR[MAXN*5];voidBuildintKintSintt) {TR[K].L=s;tr[k].r=T; TR[K].V=0; tr[k].tag=0; tr[k].sum=0; if(s==t) {Tr[k].sum=A[s]; TR[K].V=A[s]; return ; }    intMid= (s+t) >>1; Build (k<<1, S,mid); Build (k<<1|1, mid+1, T); Tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum; TR[K].V=max (tr[k<<1].v,tr[k<<1|1].v);} ll ask (intKintSintt) {    if(tr[k].l==s&&tr[k].r==t) {        returntr[k].sum; }    intMid= (TR[K].L+TR[K].R) >>1; LL ret=0; if(T<=mid) ret= Ask (k<<1, s,t); Else if(S>mid) ret= Ask (k<<1|1, s,t); Else{ret= (Ask (k<<1, S,mid) +ask (k<<1|1, mid+1, T)); } tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum; TR[K].V=max (tr[k<<1].v,tr[k<<1|1].v); returnret;}voidModifyintKintSintt,ll c) {    if(TR[K].V&LT;C)return ; if(tr[k].l==TR[K].R) {Tr[k].sum%=C; TR[K].V%=C; return ; }    intMid= (TR[K].L+TR[K].R) >>1; if(t<=mid) Modify (k<<1, S,t,c); Else if(s>mid) Modify (k<<1|1, S,t,c); Else{Modify (k<<1, s,mid,c); Modify (k<<1|1, mid+1, T,c); } tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum; TR[K].V=max (tr[k<<1].v,tr[k<<1|1].v);}voidUpdateintKintXintc) {    if(tr[k].l==x&&tr[k].r==x) {Tr[k].sum=C; TR[K].V=C; return; }    intMid= (TR[K].L+TR[K].R) >>1; if(x<=mid) Update (k<<1, X,c); Else{update (k<<1|1, X,c); } tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum; TR[K].V=max (tr[k<<1].v,tr[k<<1|1].v);}intMain () {n=read (); m=read ();  for(intI=1; i<=n;i++) {scanf ("%d",&A[i]); }build (1,1, n);intX,y;ll C;  for(intI=1; i<=m;i++) {scanf ("%d",&q); if(q==1) {scanf ("%d%d",&x,&y); printf ("%i64d\n", Ask (1, x, y)); }        Else if(q==2) {scanf ("%d%d%i64d",&x,&y,&c); Modify (1, X,y,c); }        Else{scanf ("%d%d",&x,&y); Update (1, x, y); }    }  return 0;}

Code

Codeforces Round #250 (Div. 1) D. The child and Sequence segment tree interval summation + point modification + interval modulus