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?
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.
For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.
5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3
8
5
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
49
15
23
1
9
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.
题意：区间求和 单点修改 区间取模
/* 标记没法维护 但可以发现一个性质，一个数每次取模最大也会比原来的1/2小 所以单点改，如果一个区间最大值都比模数小，就不用往后递归了。 */ #include<iostream> #include<cstdio> #include<cstring> #define N 100007 #define ll long long using namespace std; ll n,m,ans,cnt; struct tree { ll l,r,mx,sum; }tr[N<<2]; inline ll read() { ll x=0,f=1;char c=getchar(); while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } void pushup(ll k) { tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum; tr[k].mx=max(tr[k<<1].mx,tr[k<<1|1].mx); } void build(ll k,ll l,ll r) { tr[k].l=l;tr[k].r=r;tr[k].sum=tr[k].mx=0; if(l==r) { tr[k].sum=read(); tr[k].mx=tr[k].sum; return; } ll mid=l+r>>1; build(k<<1,l,mid);build(k<<1|1,mid+1,r); pushup(k); } void changemod(ll k,ll l,ll r,ll mod) { if(l>r) return; if(tr[k].mx<mod) return; if(tr[k].l==tr[k].r) { tr[k].mx%=mod; tr[k].sum=tr[k].mx; return; } ll mid=tr[k].l+tr[k].r>>1; if(r<=mid) changemod(k<<1,l,r,mod); else if(l>mid) changemod(k<<1|1,l,r,mod); else changemod(k<<1,l,mid,mod),changemod(k<<1|1,mid+1,r,mod); pushup(k); } void change(ll k,ll pos,ll x) { if(tr[k].l==tr[k].r && tr[k].l==pos) { tr[k].sum=tr[k].mx=x; return; } ll mid=tr[k].l+tr[k].r>>1; if(pos<=mid) change(k<<1,pos,x); if(pos>mid) change(k<<1|1,pos,x); pushup(k); } ll query(ll k,ll l,ll r) { if(l>r) return 0; if(tr[k].l==l && tr[k].r==r) return tr[k].sum; ll mid=tr[k].l+tr[k].r>>1; if(r<=mid) return query(k<<1,l,r); else if(l>mid) return query(k<<1|1,l,r); else return query(k<<1,l,mid)+query(k<<1|1,mid+1,r); } int main() { n=read();m=read(); build(1,1,n);ll opt,x,y,z; for(ll i=1;i<=m;i++) { opt=read(); if(opt==1) { x=read();y=read(); printf("%lld ",query(1,x,y)); } if(opt==2) { x=read();y=read();z=read(); changemod(1,x,y,z); } if(opt==3) { x=read();y=read(); change(1,x,y); } } return 0; }