寒假集训_专题四题解_B - A Simple Problem with Integers

题目

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.

题目大意

给你一个数组，要求你实现区间修改，区间查询

题解

线段树模板

代码

#include <iostream>
using namespace std;
typedef long long LL;
const int N = 1e6 + 10;
int b[N],m,n;
struct node
{
	int l,r;
	LL sum,tag;
}a[4*N];
void push_up(int now)
{
	a[now].sum = a[now<<1].sum + a[now<<1|1].sum;
}
void build(int l, int r, int now)
{
	a[now].l = l, a[now].r = r, a[now].tag = 0;
	if(l == r)a[now].sum = b[l];
	else
	{
		int mid =(l+r)>>1;
		build(l,mid,now<<1);
		build(mid+1,r,now<<1|1);
		push_up(now);
	}
}
void push_down(int x)
{
	a[x<<1].tag += a[x].tag;
	a[x<<1|1].tag += a[x].tag;
	a[x<<1].sum += a[x].tag*(a[x<<1].r-a[x<<1].l+1);
	a[x<<1|1].sum += a[x].tag*(a[x<<1|1].r-a[x<<1|1].l+1);
	a[x].tag = 0;
}
void add(int l,int r, int now, LL k)
{
	if(l<=a[now].l&&a[now].r<=r)
	{
		a[now].sum += (a[now].r-a[now].l+1)*k;
		a[now].tag += k;
		return;
	}
	if(a[now].tag)push_down(now);
	if(l<=a[now<<1].r)add(l,r,now<<1,k);
	if(a[now<<1|1].l<=r)add(l,r,now<<1|1,k);
	push_up(now);
}
LL query(int l, int r, int now)
{
	if(l<=a[now].l&&a[now].r<=r)
	{
		return a[now].sum;
	}
	if(a[now].tag)push_down(now);
	LL t = 0;
	if(l <= a[now<<1].r) t += query(l,r,now<<1);
	if(a[now<<1|1].l <= r) t += query(l,r,now<<1|1);
	return t;
}
int main()
{
	char op[4];
	cin >> n >> m;
	for(int i = 1; i <= n; i ++)
	scanf("%d",&b[i]);
	build(1,n,1);
	while(m--)
	{
		int l,r;
		cin>>op>>l>>r;
		if(op[0] == 'Q')
			cout << query(l,r,1) << endl;
		else 
		{
			LL k;
			scanf("%lld",&k);
			add(l,r,1,k);
		}
	}
	return 0;
}