区间修改，区间求和 线段树

You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

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.

#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>
#include<cstring>
#include<algorithm>
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
#define L id<<1
#define R id<<1|1
using namespace std;
const int maxn = (int)2e5 + 10;
typedef long long ll;
struct node
{
	ll lazy,sum;
	int left, right;
	void update(int x)
	{
		sum += 1ll * (right - left + 1)*x;
		lazy += x;
	}
	int mid()
	{
		return (left + right) >> 1;
	}
}tree[maxn<<2|1];
void pushup(int id)
{
	tree[id].sum = tree[L].sum + tree[R].sum;
}
void pushdown(int id)
{
	int lazyval = tree[id].lazy;
	if (lazyval)
	{
		tree[L].update(lazyval);
		tree[R].update(lazyval);
		tree[id].lazy = 0;
	}
}
void build(int id, int l, int r)
{
	tree[id].left = l, tree[id].right=r;
	tree[id].lazy = 0;
	if (l == r)
	{
		scanf("%lld",&tree[id].sum);
	}
	else
	{
		int mid = tree[id].mid();
		build(lson);
		build(rson);
		pushup(id);
	}
}
void update(int id,int l,int r,int val)
{
	if (l <= tree[id].left&&tree[id].right <= r)
		tree[id].update(val);
	else
	{
		pushdown(id);
		int mid = tree[id].mid();
		if (mid >= l)update(L, l, r, val);
		if (r > mid) update(R, l, r, val);
		pushup(id);
	}
}
ll query(int id, int l, int r)
{
	if (l <= tree[id].left&&tree[id].right <= r)
		return tree[id].sum;
	else
	{
		pushdown(id);
		ll res = 0;
		int mid = tree[id].mid();
		if (mid >= l) res+=query(L, l, r);
		if(r>mid) res+=query(R, l, r);
		pushup(id);
		return res;
	}
}
int main()
{
	int n, m,a,b,c;
	string s;
	scanf("%d%d", &n, &m);
	build(1,1,n);
	while (m--)
	{
		cin >> s;
		if (s[0] == 'Q')
		{
			scanf("%d%d",&a,&b);
			cout << query(1, a, b) << endl;
		}
		else if (s[0]=='C')
		{
			scanf("%d%d%d", &a, &b,&c);
			update(1,a,b,c);
		}
	}
	return 0;
}