POJ3468——A Simple Problem with Integers （线段树区间加值，区间求和）

Time limit

5000 ms

Case time limit

2000 ms

Memory limit

131072 kB

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.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.

代码：

#include <cstdio>
#include <cstring>

using namespace std;

const int MAXN = 100005;

long long Tree[MAXN*4];
int Data[MAXN];
long long add[MAXN*4];

void Build(int temp,int left,int right){
	if(left == right){
		Tree[temp] = Data[left];
		return ;
	}
	int mid = left + (right-left)/2;
	Build(temp<<1,left,mid);
	Build(temp<<1|1,mid+1,right);
	Tree[temp] = Tree[temp<<1]+Tree[temp<<1|1];
}

void PushDown(int temp,int left,int right){
	if(add[temp]){
		add[temp<<1] += add[temp];
		add[temp<<1|1] += add[temp];
		int mid = left + (right-left)/2;
		Tree[temp<<1] += (mid-left+1)*add[temp];
		Tree[temp<<1|1] += (right-mid)*add[temp];
		add[temp] = 0;
	}
}

void Updata(int temp,int left,int right,int ql,int qr,int value){
	if(ql<=left && qr>=right){
		add[temp] += value;
		Tree[temp] += value*(right-left+1);
		return;
	}
	
	PushDown(temp,left,right);
	
	int mid = left + (right-left)/2;
	if(ql<=mid)Updata(temp<<1,left,mid,ql,qr,value);
	if(qr>mid)Updata(temp<<1|1,mid+1,right,ql,qr,value);
	
	Tree[temp] = Tree[temp<<1]+Tree[temp<<1|1];
}

long long query(int temp,int left,int right,int ql,int qr){
	if(ql>right || qr<left)return 0;
	if(ql<=left && qr>=right)return Tree[temp];
	
	PushDown(temp,left,right);
	
	int mid = left + (right-left)/2;
	long long ans = 0;
	if(ql<=mid)ans += query(temp<<1,left,mid,ql,qr);
	if(qr>mid)ans += query(temp<<1|1,mid+1,right,ql,qr);
	return ans;
}

int main(){
	
	int N,M;
	char ch[3];
	scanf("%d %d",&N,&M);
	for(int i=1 ; i<=N ; i++){
		scanf("%d",&Data[i]);
	}
	Build(1,1,N);
	int A,B,C;
	while(M--){
		scanf("%s",ch);
		if(ch[0] == 'Q'){
			scanf("%d %d",&A,&B);
			printf("%lld
",query(1,1,N,A,B));
		}
		else if(ch[0] == 'C'){
			scanf("%d %d %d",&A,&B,&C);
			Updata(1,1,N,A,B,C);
		}
	}
	
	return 0;
}