**1080 Line tree exercise (http://codevs.cn/problem/1080/)**
time limit: 1 sspace limit: 128000 KBtitle level: Diamonds DiamondTitle Description

Description
A row of n squares, beginning with an integer in each lattice. Some questions and modifications are presented dynamically: the form of questioning is the summation of all elements in a particular sub-interval [a, b]; the modified rule is to specify a lattice x, plus or minus a specific value, a. You are now asked to make the right answer to each question. 1≤n<100000, the total number of questions and modifications m<10000.

Enter a description

Input Description
The input file first behaves as an integer n, followed by n rows n integers, representing the original integers in the lattice. followed by a positive integer m, followed by M-line, the M-query, the first integer to indicate the query code, the number 1 indicates an increase, the following two numbers x and a for the value on position x to increase a, ask the code 2 for the interval sum, followed by two integers for A and B, representing the interval between [A, a] and

Output description

Output Description
A total of M lines, 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 Tips

Data Size & Hint
1≤n≤100000, m≤10000.

#include <cstdio>Const intmaxn=100001;intA[maxn],cur=0;structtreetype{intLeft,right; intlptr,rptr; intsum;} tree[2*MAXN];voidBuildtree (intllintRR) { inttot=++cur; Tree[tot].left=ll; Tree[tot].right=RR; if(ll!=rr-1) {tree[tot].lptr=cur+1;//record the number of the left and right son (location)Buildtree (ll, (LL+RR)/2); Tree[tot].rptr=cur+1; Buildtree (LL+RR)/2, RR); Tree[tot].sum=tree[tree[tot].lptr].sum+tree[tree[tot].rptr].sum; } Elsetree[tot].sum=a[ll];}voidChangeintKintCintDelta) { if(tree[k].left==tree[k].right-1) Tree[k].sum+=Delta; Else { if(c< (tree[k].left+tree[k].right)/2) Change (Tree[k].lptr,c,delta); if(c>= (tree[k].left+tree[k].right)/2) Change (Tree[k].rptr,c,delta); Tree[k].sum=tree[tree[k].lptr].sum+tree[tree[k].rptr].sum; }}intQueryintKintllintRR) { if(Ll<=tree[k].left&&rr>=tree[k].right)returntree[k].sum; intans=0;//defined inside the function to prevent the next layer from being searched, the value of the previous layer is discarded, which is equivalent to redefining a new variable at each level if(ll< (tree[k].left+tree[k].right)/2) ans+=query (TREE[K].LPTR,LL,RR); if(Rr> (tree[k].left+tree[k].right)/2) ans+=query (TREE[K].RPTR,LL,RR); returnans;}intMain () {intN; scanf ("%d",&N); for(intI=1; i<=n;i++) scanf ("%d",&A[i]); Buildtree (1, n+1); scanf ("%d",&N); for(intI=1; i<=n;i++) { intx, Y, Z scanf ("%d%d%d",&x,&y,&z); if(x==1) Change (1, y,z); if(x==2) printf ("%d\n", Query (1, y,z+1)); } return 0;}

codevs1080 Line segment Tree Exercises