Nyoj116 soldier Kills (ii) (interval query and single-point update for segment trees)

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Soldier Kills (ii) time limit: +Ms | Memory Limit:65535KB Difficulty:5

Describe

South Generals have n soldiers, numbered 1 to N, each of these soldiers is known for their kills.

The staff is the South General's military advisers, the South General often want to know the number m to nth soldier's total kills, please help the handyman to answer the South General.

The South General's inquiry after the soldier I may also kill the Q person, after the South General asked again, need to consider the new kill number.

Input

only one set of test data
The first line is two integer n,m, where N represents the number of soldiers (1<n<1000000), and M represents the number of instructions. (1<m<100000)
The next line is n integers, and AI represents the number of soldier kills. (0<=ai<=100)
The subsequent M-line is an instruction that contains a string and two integers, first a string, and if the string query indicates that the South General has a query operation, followed by the two integer m,n, indicating the start and end of the query soldier number; If the string add is followed by the two integer I , A (1<=i<=n,1<=a<=100), which indicates that the number of new kills for the first soldier is a.

Output

for each query, output an integer r that represents the total number of kills of soldier number m to nth soldier, one row per set of outputs

Sample input

5 2 3 4 5QUERY 1 3ADD 1 2QUERY 1 3ADD 2 3QUERY 1 2QUERY 1 5

Sample output

68820

Source

[Zhang Yunzun] Original

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Zhang Yunzun

Only segment tree See also someone with a tree-like array and so will learn to learn haha.

#include <stdio.h>struct node{int Left,right;long Long Count;} tree[4*1000000];//contribution void build (int l,int r,int root) {tree[root].left=l;tree[root].right=r;if (l==r) {scanf ("%lld", &tree[root].count); return;} Else{int mid= (l+r)/2;build (l,mid,root*2); build (mid+1,r,root*2+1); tree[root].count=tree[root*2].count+tree[root* 2+1].count;}} query void search (int l,int r,int Root,long long &result) {if (tree[root].left==l&&tree[root].right==r) { Result=tree[root].count;return;} int mid= (tree[root].left+tree[root].right)/2;if (mid<l) search (L,r,root*2+1,result), else if (mid>=r) search (L, R,root*2,result); Else{long long Result1;search (L,MID,ROOT*2,RESULT1); search (mid+1,r,root*2+1,result); result= RESULT+RESULT1;}} update void update (int pos,int x,int root) {if (Tree[root].left==tree[root].right&&tree[root].right==pos) {tree[ Root].count+=x;return;} int mid= (tree[root].left+tree[root].right)/2;//Find the target soldier if (pos<=mid) update (pos,x,root*2); else if (Pos>mid) Update (pos,x,root*2+1);//When the target soldier is updatedThe parent node associated with it is updated tree[root].count=tree[root*2].count+tree[root*2+1].count;} int main () {int n,k;scanf ("%d%d", &n,&k), Build (1,n,1), for (int i=0;i<k;i++) {char str[10];int a,b;scanf ("%s%") D%d ", str,&a,&b), if (str[0]== ' Q ') {Long long Result;search (A,b,1,result);p rintf ("%lld\n ", result);} if (str[0]== ' A ') {update (a,b,1);}} return 0;}

Nyoj116 soldier Kills (ii) (interval query and single-point update for segment trees)