A Simple Problem with Integers（POJ-3468）（线段树）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 71540		Accepted: 22049
Case Time Limit: 2000MS

Description

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.

该题的关键是要高效的实现对一个区间的每个数都加上一个数，所以增加了一个数组data表示这一次加的那个数,datb表示去掉这个数的区间和。然后在sum函数里更新这两个值。

#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int DAT_SIZE = (1<<18)-1 ;
const int MAX_N = 100005;
const int MAX_Q = 100005;
int n,q,a,b,c;
char s[10];
ll data[DAT_SIZE],datb[DAT_SIZE];
void add(int a,int b,int x,int k,int l,int r){
    if(a<=l&&r<=b) data[k] += x;
    else if(l