用主席樹可以線上做，樹狀陣列只能離線。

先放個主席樹的。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<queue>
 5 #include<set>
 6 #include<string>
 7 #include<cmath>
 8 #include<cstring>
 9 #define FF first
10 #define SS second
11 #define ll long long
12
 
 #define pb push_back
13 #define pm make_pair
14 using namespace std;
15 typedef pair<int,int> PII;
16 const int MAX = 1e5 + 5;
17 struct TREE {
18     int l,r;
19     int val;
20 } tr[MAX*40];
21  
22 int tot;
23 int a[MAX],b[MAX];
24 int root[MAX];
25 int build(int l,int r) {
26     int cur = ++tot;
 
27     tr[cur].val = 0;
28     if(l == r) {
29         tr[cur].l = tr[cur].r = 0;//這一步好像沒啥用？ 
30         return cur;
31     }
32     int m = (l+r)>>1;
33     tr[cur].l = build(l,m);
34     tr[cur].r = build(m+1,r);
35     return cur;
36 }
37 void pushup(int cur) {
38     tr[cur].val = tr[tr[cur].l].val + tr[tr[cur].r].val;
 
39 } 
40 int update(int pre,int tar,int l,int r) {
41     int cur = ++tot;
42     tr[cur] = tr[pre];
43     if(l == r) {
44         tr[cur].val++;
45         return cur;
46     }
47     int m = (l+r)>>1;
48     if(tar <= m) tr[cur].l = update(tr[pre].l,tar,l,m);
49     else tr[cur].r = update(tr[pre].r,tar,m+1,r);
50     pushup(cur);
51     return cur;
52 }
53 int query(int pl,int pr,int l,int r,int H) {
54     if(l == r) return tr[pr].val - tr[pl].val;
55     int m = (l+r)>>1;
56     if(H <= m) return query(tr[pl].l,tr[pr].l,l,m,H);
57     else return tr[tr[pr].l].val - tr[tr[pl].l].val + query(tr[pl].r,tr[pr].r,m+1,r,H);
58 }
59 int main()
60 {
61     int n,m;
62     int t,iCase=0;
63     cin>>t;
64     while(t--) {
65         printf("Case %d:\n",++iCase);
66         cin>>n>>m;
67         tot=0;
68         for(int i = 1; i<=n; i++) scanf("%d",a+i),b[i] = a[i];
69         root[0] = build(1,n);
70         sort(b+1,b+n+1);
71         int LEN = unique(b+1,b+n+1) - b - 1;
72         for(int i = 1; i<=n; i++) {
73             int pos = lower_bound(b+1,b+LEN+1,a[i]) - b;
74             root[i] = update(root[i-1],pos,1,n);
75         }
76         while(m--) {
77             int l,r,H;
78             scanf("%d%d%d",&l,&r,&H);l++,r++;
79             int pos = upper_bound(b+1,b+LEN+1,H) - b - 1;
80             if(pos) printf("%d\n",query(root[l-1],root[r],1,n,pos));
81             else printf("0\n");
82         }
83     }    
84     return 0 ;
85 }

再放一個樹狀陣列的。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 using namespace std;
 7 const int maxm=1e5+5;
 8 struct Data{
 9     int num,id;
10     bool operator<(const Data a){//按數值大小排序
11         return num<a.num;
12     }
13 }a[maxm];
14 struct Query{
15     int l,r,h,id;
16     bool operator<(const Query a){//按h排序
17         return h<a.h;
18     }
19 }q[maxm];
20 int c[maxm];//BIT
21 int res[maxm];//存答案
22 int n,m;
23 int cas=1;
24 void init(){
25     memset(c,0,sizeof c);
26 }
27 int lowbit(int i){
28     return i&-i;
29 }
30 void add(int i,int x){
31     while(i<=n){
32         c[i]+=x;
33         i+=lowbit(i);
34     }
35 }
36 int ask(int i){
37     int ans=0;
38     while(i){
39         ans+=c[i];
40         i-=lowbit(i);
41     }
42     return ans;
43 }
44 void input(){
45     scanf("%d%d",&n,&m);
46     for(int i=1;i<=n;i++){
47         scanf("%d",&a[i].num);
48         a[i].id=i;
49     }
50     for(int i=1;i<=m;i++){
51         scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].h);
52         q[i].l++,q[i].r++;//題目區間從0開始，加1變為從1開始
53         q[i].id=i;
54     }
55     sort(a+1,a+1+n);
56     sort(q+1,q+1+m);
57 }
58 void solve(){
59     printf("Case %d:\n",cas++);
60     int last=1;
61     for(int i=1;i<=m;i++){
62         while(q[i].h>=a[last].num&&last<=n){//把數值比h小的插入
63             add(a[last].id,1);//插入的是編號
64             last++;
65         }
66         res[q[i].id]=ask(q[i].r)-ask(q[i].l-1);
67     }
68     for(int i=1;i<=m;i++){
69         printf("%d\n",res[i]);
70     }
71 }
72 int main(){
73     int T;
74     scanf("%d",&T);
75     while(T--){
76         init();
77         input();
78         solve();
79     }
80     return 0;
81 }

HDU 4417，板題。