P3368 [TEMPLATE] tree array 2 (tree array maintains the differential sequence), p3368 tree

P3368 [TEMPLATE] tree array 2 (tree array maintains the differential sequence), p3368 tree
Description

For example, if you know a sequence, you need to perform the following two operations:

1. Add x to each number in a certain range

2. Obtain the sum of a certain number

Input/Output Format

Input Format:

 

The first line contains two integers N and M, indicating the number of numbers in the series and the total number of operations respectively.

The second row contains N integers separated by spaces. the I-th digit indicates the initial value of the I-th Column.

Each row in the next M row contains 2 or 4 integers, indicating an operation, as shown below:

Operation 1: Format: 1 x y k meaning: Add k to each number in the range [x, y]

Operation 2: Format: 2 x meaning: output the value of number x

 

Output Format:

 

The output contains several integer rows, that is, the result of all operation 2.

 

Input and Output sample input sample #1:

5 51 5 4 2 31 2 4 22 31 1 5 -11 3 5 72 4

Output sample #1:

610

Description

Time-Space limit: 1000 ms, 128 M

Data scale:

For 30% of data: N <= 8, M <= 10

For 70% of data: N <= 10000, M <= 10000

For 100% of data: N <= 500000, M <= 500000

Example:

Therefore, the output result is 6 and 10.

 

 

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 using namespace std; 6 int lowbit (int x) 7 {return x &-x;} 8 void read (int & n) 9 {10 char c = '+'; int x = 0; bool flag = 0; 11 while (c <'0' | c> '9') 12 {c = getchar (); if (c = '-') flag = 1 ;} 13 while (c> = '0' & c <= '9') 14 {x = x * 10 + (c-48), c = getchar ();} 15 flag = 1? N =-x: n = x; 16} 17 const int MAXN = 500001; 18 int c [MAXN], n, m, p, x, y, z, pre; 19 void add (int p, int v) 20 {21 while (p <= n) 22 {23 c [p] + = v; 24 p + = lowbit (p ); 25} 26} 27 int ask (int p) 28 {29 int ans = 0; 30 while (p> 0) 31 {32 ans + = c [p]; 33 p-= lowbit (p); 34} 35 return ans; 36} 37 int main () 38 {39 read (n); read (m ); 40 for (int I = 1; I <= n; I ++) 41 {42 read (p); 43 add (I, p-pre); 44 pre = p; 45} 46 47 while (m --) 48 {49 read (p); 50 if (p = 1) // The interval is added with 51 {52 read (x ); read (y); read (z); 53 add (x, z); 54 add (y + 1,-z ); 55} 56 else // Single Point query 57 {58 read (x); 59 printf ("% d \ n", ask (x); 60} 61} 62 return 0; 63}