A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K
Case Time Limit: 2000MS

Description

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

n个数，q个操作
Q a b 查询a-b的和
C a b c a-b都加c

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#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
typedef long long ll;
struct node
{
	ll num;
	ll flag;
}tre[5000000 + 5];
void build(int t, int l, int r)
{
	tre[t].flag = 0;
	if (l == r)
	{
		scanf("%lld", &tre[t].num);
		return;
	}
	int mid = l + r >> 1;
	build(t << 1, l, mid);
	build(t  << 1 | 1, mid + 1, r);
	tre[t].num = tre[t << 1].num + tre[t << 1 | 1].num;
}
ll ans;
int getlen(int l, int x, int r, int y)
{
	return min(r, y) - max(l, x) + 1;
}
void update(int t, int l, int r, int x, int y, int k)
{
	if (l >= x&&r <= y)
	{
		tre[t].flag += k;
		return;
	}
	tre[t].num += (getlen(l, x, r, y)*k);
	int mid = l + r >> 1;
	if (y <= mid)
		update(t << 1, l, mid, x, y, k);
	else if (x > mid)
		update(t << 1 | 1, mid + 1, r, x, y, k);
	else
	{
		update(t << 1, l, mid, x, y, k);
		update(t << 1 | 1, mid + 1, r, x, y, k);
	}
}
void query(int t, int l, int r, int x, int y, ll f)
{
	if (l >= x&&r <= y)
	{
		ans += tre[t].num;
		ans += (f+tre[t].flag)*getlen(l,x,r,y);
		return;
	}
	int mid = l + r >> 1;
	if (y <= mid)
		query(t << 1, l, mid, x, y, f + tre[t].flag);
	else if (x > mid)
		query(t << 1 | 1, mid + 1, r, x, y, f + tre[t].flag);
	else
	{
		query(t << 1, l, mid, x, y, f + tre[t].flag);
		query(t << 1 | 1, mid + 1, r, x, y, f + tre[t].flag);
	}
}
int main()
{
	int n, q;
	scanf("%d%d", &n, &q);
	build(1, 1, n);
	while (q--)
	{
		char s[5];
		int a, b, c;
		scanf("%s%d%d", s, &a, &b);
		if (s[0] == 'Q')
		{
			ans = 0;
			query(1, 1, n, a, b, 0);
			printf("%lld\n", ans);
		}
		else
		{
			scanf("%d", &c);
			update(1, 1, n, a, b, c);
		}
	}
}