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 The sums may exceed the range of 32-bit integers.

程式碼:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 100050;
ll digit[maxn];
struct node
{
	ll sum,lazy;
	int left,right;
}tree[maxn*4];
//線段樹的建立 
void build(int k,int l,int r)
{
	tree[k].left = l;
	tree[k].right = r;
	tree[k].lazy = 0;//懶惰標記為 0 因為是區間更新所以容易超時這個時候就應該lazy標記 
	if(l==r)
	{
		tree[k].sum = digit[l];
		return ;
	}
	int mid = (l+r)/2;
	build(k<<1,l,mid);
	build(k<<1|1,mid+1,r);
	tree[k].sum = tree[k<<1].sum + tree[k<<1|1].sum;
} 
//釋放懶惰標記
void pushlazy(int k)
{
	if(tree[k].lazy)
	{
		tree[k<<1].lazy += tree[k].lazy; 
		tree[k<<1|1].lazy += tree[k].lazy;
		tree[k<<1].sum +=tree[k].lazy*(tree[k<<1].right-tree[k<<1].left+1); //這邊和下面那一行要注意的就是是！！tree[k].lazy*
		tree[k<<1|1].sum +=tree[k].lazy*(tree[k<<1|1].right-tree[k<<1|1].left+1);
		tree[k].lazy = 0;
	} 
} 

//線段樹的更新
void update(int k,int l,int r,ll val)
{
	if(tree[k].left == l && tree[k].right == r)
	{
		tree[k].lazy +=val; //因為下面的其他節點都還沒有加上他們所應該的值所以不能為0 
		tree[k].sum += val * (r-l+1); //這一步是因為要加上這個的值
		return ;	
	}	
	if(tree[k].left == tree[k].right) return; //因為是葉子節點所以說不用再向下釋放了直接可以return
	pushlazy(k);//釋放lazy標記
	int mid = (tree[k].left + tree[k].right)>>1;
	if(r<=mid)
		update(k<<1,l,r,val);
	else if(l>mid)
		update(k<<1|1,l,r,val);
	else 
	{
		update(k<<1,l,mid,val);
		update(k<<1|1,mid+1,r,val);
	}
	tree[k].sum  = tree[k<<1].sum + tree[k<<1|1].sum;//這步要記得加
	return;
}
ll query(int k,int l,int r)
{
	if(tree[k].left == l && r == tree[k].right)
	{
		return tree[k].sum;
	}
	pushlazy(k);
	ll ans = 0;
	int mid = (tree[k].left + tree[k].right)>>1;
	if(r<=mid)
		ans+=query(k<<1,l,r);
	else if(l>mid)
		ans+=query(k<<1|1,l,r);
	else 
	{
		ans+=query(k<<1,l,mid);
		ans+=query(k<<1|1,mid+1,r);
	}
	return ans;
} 
int main()
{
	int n,q,x,y;ll z;
	scanf("%d%d",&n,&q);
	for(int i = 1;i<=n;i++)
		scanf("%lld",&digit[i]);
	build(1,1,n);
	while(q--)
	{
		char s[5];
		scanf("%s",s);
		if(s[0] == 'Q')
		{
			scanf("%d%d",&x,&y);
			printf("%lld\n",query(1,x,y));
		}else 
		{
			scanf("%d%d%lld",&x,&y,&z);
			update(1,x,y,z);
		}
	}
 	return 0;
} 
 

query其他寫法

ll query(int k,int l,int r)
{
	if(tree[k].left >= l &&  tree[k].right <=r)
	{
		return tree[k].sum;
	}
	pushlazy(k);
	ll ans = 0;
	int mid = (tree[k].left + tree[k].right)>>1;
	if(l<=mid)
		ans+=query(k<<1,l,r);
	if(r>mid)
		ans+=query(k<<1|1,l,r);
	return ans;
}