HDU1166线段树基础题

线段树从零学起，先练习一个最简单的（虽说最简单的我也搞了一天）

////  test.cpp//  HDU1166////  Created by dan on 16/9/29.//  Copyright © 2016年 dan. All rights reserved.//#include <cstring>#include <iostream>#define MAXN 50005using namespace std;int sav[MAXN];int sum[MAXN * 4];void pushUp(int i){    sum[i] = sum[i * 2] + sum[i * 2 + 1];}void build(int index, int l, int r){    if (l == r)    {        sum[index] = sav[l];//叶子节点        return;    }    else{        build(index * 2, l , (l + r) / 2); //递归的对他的儿子结点构造，由于是完美二叉树，孩子节点        //为index << 1与 index <<1 | 1,相应的，令 m = (r +l) / 2;        //对应区间为(l, m / 2), (m + 1, r);最后更新父亲节点的值，无论是最小值还是求和        build(index * 2 + 1, (l + r) / 2 + 1, r);        pushUp(index);    }}void update(int bianhao, int addnum, int l, int r, int cunchu){    if (l == r)        sum[cunchu] += addnum;    else{        int m = (l + r) / 2;        if (bianhao <= m){//这里的目的是找到bianhao对应的数组位置.            update(bianhao, addnum, l, m, 2 * cunchu);        }        else            update(bianhao, addnum, m + 1, r , 2 * cunchu + 1);        pushUp(cunchu);    }}int query(int begin, int end, int l, int r, int index){    if (l >= begin && r <= end){        return sum[index];    }    int m = (l + r) / 2;    int ans = 0;    if (begin <= m){        ans += query(begin, end, l, m, index * 2);        //由于这里是递归求和，因此左右孩子都要考虑到    }    if ( m < end ){        ans += query(begin, end, m + 1, r, index * 2 + 1);    }    return ans;}int main(){    int T, n, a, b;    scanf("%d",&T);    for( int i = 1; i <= T; ++i )    {        printf("Case %d:\n",i);        scanf("%d",&n);        for (int i = 1; i <= n; i++){            scanf("%d", &sav[i]);        }        build(1,1, n);        char op[10];        while( scanf("%s",op) &&op[0]!='E' )        {            scanf("%d %d", &a, &b);            if(op[0] == 'Q')                printf("%d\n",query(a,b,1,n,1));            else if(op[0] == 'S')                update(a,-b,1,n,1);            else                update(a,b,1,n,1);        }    }      return 0;}