HDU 1166 线段树的单点更新 区间求和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define N 50005
using namespace std;

int num
;
struct Tree
{
int l;
int r;
int sum;
}tree[N * 4];

void build(int root,int l,int r)///root为节点 此处为根节点 1 l，r为线段的左端点和右端点
{
tree[root].l = l;
tree[root].r = r;
if(tree[root].l == tree[root].r)
{
tree[root].sum = num[l];
return ;
}
int mid = (l + r) / 2;
build(root << 1,l ,mid);
build(root << 1|1,mid+1,r);
tree[root].sum = tree[root << 1].sum + tree[root << 1|1].sum;///此处灵活应用根据题意不同
}
void update(int root,int pos,int val)///感觉此处类似回溯 先找到最低层 更新所要求的值然后一层一层的递归上来
{
if(tree[root].l == tree[root].r)///找到最底层的数值 更新
{
tree[root].sum = val;
return ;
}
int mid=(tree[root].l + tree[root].r) / 2;
if(pos <= mid)
update(root << 1,pos,val);
else
update(root << 1|1,pos,val);
tree[root].sum = tree[root << 1].sum + tree[root << 1|1].sum;///更新
}
int query(int root,int L,int R)///求和逐层递归就好
{
if(L <= tree[root].l && R >= tree[root].r)
return tree[root].sum;

int mid = (tree[root].l + tree[root].r) / 2,ret = 0;

if(L <= mid ) ret += query(root << 1,L,R);
if(R > mid) ret += query(root << 1|1,L,R);
return ret;
}
int main()
{
int ca,cas = 1,n,Q,a,b;
char str[10];
scanf("%d",&ca);
while(ca--)
{
scanf("%d",&n);
for(int i = 1;i <= n;i ++)
scanf("%d",&num[i]);
build(1,1,N);
printf("Case %d:\n",cas++);
while(scanf("%s",str),strcmp(str,"End"))
{
scanf("%d%d",&a,&b);
if(strcmp(str,"Query") == 0)
{
if(a > b) swap(a , b);
printf("%d\n",query(1,a,b));
}
else if(strcmp(str,"Add") == 0)
{
num[a] = num[a] + b;
update(1,a,num[a]);
}
else if(strcmp(str,"Sub") == 0)
{
num[a] = num[a] - b;
update(1,a,num[a]);
}
}
}
return 0;
}