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Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
C a b c means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
Q a b means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

Source

题目大意：C情况时a-b区间内每个数字添加c，Q情况时查询区间内值的和。

解题思路：模板线段树成段更新问题。

代码如下：

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#include 
  
   
#include 
   
     #define inf 1e12 #define ll long long const ll maxn=1000010; ll n,m,a,b,c; ll sum[maxn],nd[maxn]; char s[2]; void build(ll l,ll r,ll root) { ll mid=(l+r)/2; if(l==r) { scanf(%lld,&sum[root]); return; } build(l,mid,root*2); build(mid+1,r,root*2+1); sum[root]=sum[root*2]+sum[root*2+1]; } void pushdown(ll l,ll root) { if(nd[root]!=0) { nd[root*2]+=nd[root]; nd[root*2+1]+=nd[root]; sum[root*2]+=((l-l/2)*nd[root]); sum[root*2+1]+=((l/2)*nd[root]); nd[root]=0; } } ll query(ll l,ll r,ll root) { if(l>=a&&r<=b) return sum[root]; pushdown(r-l+1,root); ll mid=(l+r)/2,ans=0; if(a<=mid) ans+=query(l,mid,root*2); if(b>mid) ans+=query(mid+1,r,root*2+1); return ans; } void update(ll l,ll r,ll root) { if(l>b||r
    
     =a&&r<=b) { nd[root]+=c; sum[root]+=(len*c); return ; } pushdown(len,root); int mid=(l+r)/2; if(a<=mid) update(l,mid,root*2); if(b>mid) update(mid+1,r,root*2+1); sum[root]=sum[root*2]+sum[root*2+1]; } int main(void) { memset(nd,0,sizeof(nd)); scanf(%lld%lld,&n,&m); build(1,n,1); for(ll i=0;i
     
      
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