題干：

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?moperations. 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.

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.

題目大意：

給一個序列
支持3種操作
1 u v 對于所有i u<=i<=v，輸出a[i]的和
2 u v t 對于所有i u<=i<=v a[i]=a[i]%t
3 u v 表示a[u]=v（將v賦值給a[u]）
n,q<=1e5 a[i],t,v<=1e9

解題報告：

? ? 這題需要維護最大值的，不能直接判斷是否sum=0了就返回。因為我操作可能給的模數很大，如果我給1e5次這個大模數操作，那就n^2logn了，肯定炸，但是用判斷最大值的的方法就可以取消一大部分操作使復雜度降回來。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair using namespace std; const int MAX = 2e5 + 5; struct TREE {int l,r;ll sum;ll maxx; } tree[MAX<<2]; ll a[MAX]; int n,m; void pushup(int cur) {tree[cur].sum = tree[cur*2].sum + tree[cur*2+1].sum;tree[cur].maxx = max(tree[cur*2].maxx,tree[cur*2+1].maxx); } void build(int l,int r,int cur) {tree[cur].l = l,tree[cur].r = r;if(l == r) {tree[cur].sum = tree[cur].maxx = a[r];return ;}int m = (l+r)>>1;build(l,m,cur*2);build(m+1,r,cur*2+1);pushup(cur); } void update(int pl,int pr,ll val,int cur) {if(tree[cur].maxx < val) return ;if(tree[cur].l == tree[cur].r) {tree[cur].sum %=val;tree[cur].maxx %= val;return ;}if(pl <= tree[cur*2].r) update(pl,pr,val,cur*2);if(pr >= tree[cur*2+1].l) update(pl,pr,val,cur*2+1);pushup(cur); } void update2(int tar,ll val,int cur) {if(tree[cur].l == tree[cur].r) {tree[cur].sum = tree[cur].maxx = val;return ;}if(tar <= tree[cur*2].r) update2(tar,val,cur*2);if(tar >= tree[cur*2+1].l) update2(tar,val,cur*2+1);pushup(cur); } ll qSum(int pl,int pr,int cur) {if(pl <= tree[cur].l && pr >= tree[cur].r) return tree[cur].sum;ll res = 0;if(pl <= tree[cur*2].r) res += qSum(pl,pr,cur*2);if(pr >= tree[cur*2+1].l) res += qSum(pl,pr,cur*2+1); return res; } int main() {cin>>n>>m;for(int i = 1; i<=n; i++) cin>>a[i];build(1,n,1);while(m--) {int op,u,v;ll t;scanf("%d",&op);if(op == 1) {scanf("%d%d",&u,&v);printf("%lld\n",qSum(u,v,1));}if(op == 2) {scanf("%d%d%lld",&u,&v,&t);if(u>v) swap(u,v);update(u,v,t,1);}if(op == 3) {scanf("%d%lld",&u,&t);update2(u,t,1);}}return 0 ; } /* 16:25 - 16:38 */

?

總結

以上是生活随笔為你收集整理的【CodeForces - 438D】The Child and Sequence（线段树区间取模操作）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。