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#include #include #define N 100007#define ll long longusing 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 >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\n",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;}