1080 line segment tree exercise single-point modification and interval query, 1080 Line Segments

1080 line segment tree exercise single-point modification and interval query, 1080 Line Segments
1080 line segment tree exercises

 

Time Limit: 1 s space limit: 128000 KB title level: Diamond Title Description Description

There are N squares in a row. At the beginning, each grid contains an integer. Now, I want to dynamically raise some questions and modify them: the form of a question is to find the sum of all elements in a specific subinterval [a, B; the modified rule is to specify A lattice x and add or subtract A specific value. Now you are required to answer each question correctly. 1 ≤ N <100000, total number of questions and modifications m <10000.

Input description Input Description

The first row of the input file is an integer N, followed by n integers in n rows, representing the original integer in the grid. The next positive integer is m, followed by m, which indicates m queries. The first integer indicates the query code and the query code 1 indicates the increase, the two numbers x and A indicate that A is added to the value at the position X. The query Code 2 indicates the sum of the intervals. The two integers a and B indicate that [, b.

Output description Output Description

A total of m rows, each integer

Sample Input Sample Input

6

4

5

6

2

1

3

4

1 3 5

2 1 4

1 1 9

2 2 6

Sample output Sample Output

22

22

Data range and prompt Data Size & Hint

1 ≤ N ≤ 100000, m ≤ 10000.

CATEGORY tag Tags click here to expand

After changing the line segment tree style, I will write it in this style.

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 using namespace std; 6 const int MAXN = 100001; 7 int n, m; 8 int ans = 0; 9 struct node10 {11 int l, r, w, f; 12} tree [MAXN * 4]; 13 void updata (int k) 14 {15 tree [k]. w = tree [k * 2]. w + tree [k * 2 + 1]. w; 16} 17 void build (int k, int ll, int rr) 18 {19 tree [k]. l = ll; tree [k]. r = rr; 20 if (tree [k]. l = tree [k]. r) 21 {22 scanf ("% d", & tree [k]. w); 23 return; 24} 25 int m = (ll + rr)/2; 26 build (k * 2, ll, m ); 27 build (k * 2 + 1, m + 1, rr); 28 updata (k); 29} 30 void point_change (int k, int p, int v) 31 {32 if (tree [k]. l = tree [k]. r) 33 {34 tree [k]. w + = v; 35 return; 36} 37 int m = (tree [k]. l + tree [k]. r)/2; 38 if (p <= m) point_change (k * 2, p, v); 39 else point_change (k * 2 + 1, p, v ); 40 updata (k); 41} 42 void ask_interval (int k, int ll, int rr) 43 {44 if (ll <= tree [k]. l & rr> = tree [k]. r) 45 {46 ans + = tree [k]. w; 47 return; 48} 49 int m = (tree [k]. l + tree [k]. r)/2; 50 if (ll <= m) 51 ask_interval (k * 2, ll, rr); 52 if (rr> m) 53 ask_interval (k * 2 + 1, ll, rr); 54} 55 int main () 56 {57 scanf ("% d", & n); 58 build (1,1, n); 59 scanf ("% d", & m); 60 for (int I = 1; I <= m; I ++) 61 {62 int how, p, v, x, y; 63 scanf ("% d", & how); 64 if (how = 1) // Add 65 {66 scanf ("% d", & p, & v); 67 point_change (1, p, v ); 68} 69 else // sum of the intervals 70 {71 scanf ("% d", & x, & y); 72 ans = 0; 73 ask_interval (1, x, y); 74 printf ("% d \ n", ans); 75} 76} 77 return 0; 78}