题意：询问区间[ L , R] 中比 H 小的数的个数，注意题目中给的区间是从0下标开始的。

思路：用upper_bound跑出 H 的相对大小，去查找 L-1 和 R 时间戳小于H的的改变数量

注意点：如果找到的H相对大小为0，就不用去寻找（否则会T），可以直接输出0即可。

///#include<bits/stdc++.h>
///#include<unordered_map>
///#include<unordered_set>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<bitset>
#include<set>
#include<stack>
#include<map>
#include<list>
#include<new>
#include<vector>
#define MT(a,b) memset(a,b,sizeof(a));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double pai=acos(-1.0);
const double E=2.718281828459;
const ll mod=1e9+7;
const int INF=0x3f3f3f3f;
const int maxn=1e5+10;


int num[100005];
int ranks[100005];
int counts;

struct node
{
    int l;
    int r;
    int cnt;
} point[4000005];

int root[100005];


void pushup(int sign)
{
    point[sign].cnt=point[point[sign].l].cnt+point[point[sign].r].cnt;
}

int buildtree(int l,int r)
{
    int s=++counts;
    point[s].cnt=0;
    if(l==r)
        return s;
    int mid=(l+r)>>1;
    point[s].l=buildtree(l,mid);
    point[s].r=buildtree(mid+1,r);
    return s;
}

int addtree(int left,int l,int r,int pos)
{
    int s=++counts;
    point[s]=point[left];
    if(l==r)
    {
        point[s].cnt++;
        return s;
    }
    int mid=(l+r)>>1;
    if(pos<=mid)
        point[s].l=addtree(point[left].l,l,mid,pos);
    else
        point[s].r=addtree(point[left].r,mid+1,r,pos);
    pushup(s);
    return s;
}

int query(int l,int r,int i,int j,int k)
{
    if(r<=k)
        return point[j].cnt-point[i].cnt;
    int mid=(l+r)>>1;
    if(k<=mid)
        return query(l,mid,point[i].l,point[j].l,k);
    else
        return point[point[j].l].cnt-point[point[i].l].cnt+query(mid+1,r,point[i].r,point[j].r,k);
}
int main()
{
    int t,n,q,pos;
    int l,r,k;
    scanf("%d",&t);
    for(int u=1; u<=t; u++)
    {
        scanf("%d %d",&n,&q);
        counts=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&num[i]);
            ranks[i]=num[i];
        }
        sort(ranks+1,ranks+1+n);
        int d=unique(ranks+1,ranks+1+n)-(ranks+1);
        root[0]=buildtree(1,d);
        for(int i=1; i<=n; i++)
        {
            pos=lower_bound(ranks+1,ranks+1+d,num[i])-ranks;
            root[i]=addtree(root[i-1],1,d,pos);
        }
        printf("Case %d:\n",u);
        while(q--)
        {
            scanf("%d %d %d",&l,&r,&k);
            l++;
            r++;
            k=upper_bound(ranks+1,ranks+1+d,k)-ranks-1;
            if(k==0)
                printf("0\n");
            else
                printf("%d\n",query(1,d,root[l-1],root[r],k));
        }
    }
    return 0;
}