hdu 1166 线段树（单点增减 区间求和）

Sample Input
1
10
1 2 3 4 5 6 7 8 9 10
Query 1 3
Add 3 6
Query 2 7
Sub 10 2
Add 6 3
Query 3 10
End

Sample Output
Case 1:
6
33
59

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <cmath>
# include <queue>
# define LL long long
using namespace std ;

const int maxn = 50010;

int sum[maxn<<2] ; //结点开4倍

void PushUP(int rt) //更新到父节点
{
    sum[rt] = sum[rt * 2] + sum[rt * 2 + 1] ; //rt 为当前结点
}

void build(int l , int r , int rt) //构建线段树
{
    if (l == r)
    {
        scanf("%d" , &sum[rt]) ;
        return ;
    }
    int m = (l + r) / 2 ;
    build(l , m , rt * 2) ;
    build(m + 1 , r , rt * 2 +1) ;
    PushUP(rt) ;
}

void updata(int p , int add , int l , int r , int rt)  //单点增减
{
    if (l == r)
    {
        sum[rt] += add ;
        return ;
    }
    int m = (l + r) / 2 ;
    if (p <= m)
       updata(p , add , l , m , rt * 2) ;
    else
       updata(p , add , m + 1 , r , rt * 2 + 1) ;
    PushUP(rt) ;
}

int query(int L , int R , int l , int r , int rt)  //区间求和
{
    if (L <= l && r <= R)
        return sum[rt] ;
    int m = (l + r) / 2 ;
    int ret = 0 ;
    if (L <= m)
       ret += query(L , R , l , m , rt * 2) ;
    if (R > m)
       ret += query(L , R , m + 1 , r , rt * 2 + 1) ;
    return ret ;
}

int main ()
{
    //freopen("in.txt","r",stdin) ;
    int T  , n;
    int Case = 0 ;
    scanf("%d" , &T) ;
    while (T--)
    {
        scanf("%d" , &n) ;
        build(1 , n , 1) ;
        char op[10] ;
        Case++ ;
        printf("Case %d:
" , Case) ;
        while (scanf("%s" , op))
        {
            if (op[0] == 'E') //结束
               break ;
            int a , b ;
            scanf("%d %d" , &a , &b) ;
            if (op[0] == 'Q') //求a,b区间的和
               printf("%d
", query(a , b , 1 , n , 1)) ;
            else if (op[0] == 'S')  //将第a个地方减少b
               updata(a , -b , 1 , n , 1) ;
            else if (op[0] == 'A')   //将第a个地方增加b
               updata(a , b , 1 , n , 1) ;
        }
    }


    return 0 ;
}