题解参考：
语法参考：
线段树参考：
1 #include2 #include 3 #include 4 using namespace std; 5 const int N=4*5*1e4+10; 6 struct node 7 { 8 int l,r,w,f;//w该区间和，f懒惰标记 9 }tree[N];10 int x,y,ans;11 void build(int k,int ll,int rr)//建树，k节点序号,ll左端点，rr右端点12 {13 tree[k].l=ll,tree[k].r=rr;14 if(tree[k].l==tree[k].r)15 {16 scanf("%d",&tree[k].w);17 return;18 }19 int m=(ll+rr)/2;20 build(k*2,ll,m);21 build(k*2+1,m+1,rr);22 tree[k].w=tree[k*2].w+tree[k*2+1].w;23 }24 void down(int k)//标记下传25 {26 tree[k*2].f+=tree[k].f;27 tree[k*2+1].f+=tree[k].f;28 tree[k*2].w+=tree[k].f*(tree[k*2].r-tree[k*2].l+1);29 tree[k*2+1].w+=tree[k].f*(tree[k*2+1].r-tree[k*2+1].l+1);30 tree[k].f=0;31 }32 void change_point(int k)//单点修改33 {34 if(tree[k].l==tree[k].r)35 {36 tree[k].w+=y;37 return;38 }39 if(tree[k].f) down(k);40 int m=(tree[k].l+tree[k].r)/2;41 if(x<=m) change_point(k*2);42 else change_point(k*2+1);43 tree[k].w=tree[k*2].w+tree[k*2+1].w;44 }45 void ask_interval(int k)//区间查询46 {47 if(tree[k].l>=x&&tree[k].r<=y)48 {49 ans+=tree[k].w;50 return;51 }52 if(tree[k].f) down(k);53 int m=(tree[k].l+tree[k].r)/2;54 if(x<=m) ask_interval(k*2);55 if(y>m) ask_interval(k*2+1);56 }57 int main()58 {59 int t;60 // freopen("in.txt","r",stdin);61 scanf("%d",&t);//用cin,cout会超时！62 63 for (int ca=1;ca<=t;ca++)//ca案例数64 {65 printf("Case %d:\n",ca);66 int n;67 scanf("%d",&n);68 build(1,1,n);69 while (1)70 {71 char ts[12];72 scanf("%s",ts);73 if (strcmp(ts,"End")==0)//这个判断要放在x,y的输入前！74 {75 break;76 }77 scanf("%d %d",&x,&y);78 ans=0;//记得清零！79 if (strcmp(ts,"Query")==0)80 {81 ask_interval(1);82 printf("%d\n",ans);83 }84 if (strcmp(ts,"Add")==0)85 {86 change_point(1);87 }88 if (strcmp(ts,"Sub")==0)89 {90 y=-y;91 change_point(1);92 }93 }94 }95 96 return 0;97 }