线段树hdu1166

线段树节点的左儿子[a,(a+b)/2]右儿子[(a+b)/2,b]

这个很重要

把线段树当成完全二叉树算，开到3*N就可以，但最好是4*N

#include<bits/stdc++.h>
using namespace std;
int a[50005];
struct {
    int sum;
    int left,right;
}b[50005*4];
void build(int left,int right,int i){
    int mid;
    b[i].left=left;b[i].right=right;
    if(left==right) {
        b[i].sum=a[left];
        return ;
    }
    mid = (left+right)/2;
    build(left,mid,2*i);
    build(mid+1,right,2*i+1);
    b[i].sum=b[2*i].sum+b[2*i+1].sum;
}
int Query(int left,int right,int i){
    int mid;
    if(b[i].left==left&&b[i].right==right) return b[i].sum;
    mid=(b[i].left + b[i].right)/2;
    if(right <= mid) return Query(left,right,2*i);
    else if(left > mid ) return Query(left,right,2*i+1);
    else return Query(left,mid,2*i) + Query(mid+1,right,2*i+1);
}
void Add(int id,int num,int i){
    if(b[i].left == b[i].right){
        b[i].sum = b[i].sum + num; return ;
    }
    else {
        b[i].sum = b[i].sum + num;
        if(id<=b[i*2].right) Add(id,num,2*i);
        else Add(id,num,2*i+1);
    }
}
int main(){
    int t,kase=1;
    scanf("%d",&t);
    while(t--){
        memset(b,0,sizeof(b));
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        build(1,n,1);
        string s;int x,y;
        printf("Case %d:\n",kase++);
        while(cin>>s){
            if(s[0]=='E') break;
            scanf("%d%d",&x,&y);
            if(s[0]=='Q'){
                printf("%d\n",Query(x,y,1));
            }
            if(s[0]=='A'){
                Add(x,y,1);
            }
            if(s[0]=='S') Add(x,-y,1);

        }
    }
}