Nyoj116 soldier kills (ii) line segment tree Single point update

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

688

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace STD; #define MAXN 1000001#define inf 0x3f3f3f3fint ll[maxn<<2],rr[maxn<<2],sum[maxn<<2];int A[MAXN]    ; void build (int l,int r,int i) {ll[i]=l;    Rr[i]=r;        if (Ll[i]==rr[i]) {sum[i]=a[l];    return;    } int m= (L+R) >>1,ls=i<<1,rs=ls|1;    Build (L,m,ls);    Build (M+1,R,RS); SUM[I]=SUM[LS]+SUM[RS];}        void update (int k,int V,int i) {if (Ll[i]==rr[i]) {sum[i]+=v;    Return    } int m= (Ll[i]+rr[i]) >>1,ls=i<<1,rs=ls|1;    if (k<=m) {update (K,V,LS);    } else{Update (K,V,RS); } Sum[i]=sum[ls]+sum[rs];}    int finds (int l,int r,int i) {if (ll[i]==l&&rr[i]==r) {return sum[i];    } int m= (Ll[i]+rr[i]) >>1,ls=i<<1,rs=ls|1;    if (r<=m) {return finds (L,R,LS);    } else if (l>m) {return finds (L,R,RS); } else{return finds (L,m,ls) +fiNDS (M+1,R,RS);    }}int Main () {int n,m;    int x, y;    int a,i;    Char q[10];    Freopen ("In.txt", "R", stdin);        while (~SCANF ("%d%d", &n,&m)) {for (int i=1;i<=n;i++) {scanf ("%d", &a[i]);        } build (1,n,1);            for (int i=1;i<=m;i++) {scanf ("%s", q);                if (!strcmp (q, "QUERY")) {scanf ("%d%d", &x,&y);            printf ("%d\n", Finds (x,y,1));                } else {scanf ("%d%d", &a,&i);            Update (a,i,1); }        }    }}

Nyoj116 soldier kills (ii) line segment tree Single point update