Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.

Case 1:

4

0

0

3

1

2

0

1

5

1
分析：离线离散化后，按时间插入主席树，最后求下区间个数和即可；
代码：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#include <list>

#define rep(i,m,n) for(i=m;i<=n;i++)

#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)

#define mod 1000000007

#define inf 0x3f3f3f3f

#define vi vector<int>

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define ll long long

#define pi acos(-1.0)

#define pii pair<int,int>

#define Lson L, mid, rt<<1

#define Rson mid+1, R, rt<<1|1

const int maxn=1e5+;

using namespace std;

ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

int n,m,k,t,a[maxn],b[maxn*],s[maxn*],ls[maxn*],rs[maxn*],root[maxn*],sz;

inline ll read()

{

    ll x=;int f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

struct node

{

    int x,y,z;

    node(){}

    node(int _x,int _y,int _z):x(_x),y(_y),z(_z){}

}op[maxn];

void insert(int l,int r,int x,int &y,int v)

{

    y=++sz;

    s[y]=s[x]+;

    if(l==r)return;

    ls[y]=ls[x],rs[y]=rs[x];

    int mid=l+r>>;

    if(v<=mid)insert(l,mid,ls[x],ls[y],v);

    else insert(mid+,r,rs[x],rs[y],v);

}

int query(int l,int r,int L,int R,int x,int y)

{

    if(l==L&&r==R)return s[y]-s[x];

    int mid=L+R>>;

    if(r<=mid)return query(l,r,L,mid,ls[x],ls[y]);

    else if(l>mid)return query(l,r,mid+,R,rs[x],rs[y]);

    else return query(l,mid,L,mid,ls[x],ls[y])+query(mid+,r,mid+,R,rs[x],rs[y]);

}

int main()

{

    int i,j;

    scanf("%d",&t);

    while(t--)

    {

        sz=;

        scanf("%d%d",&n,&m);

        rep(i,,n)a[i]=read(),b[i]=a[i];

        rep(i,,m)

        {

            int c,d,e;

            c=read(),d=read(),e=read();

            c++,d++;

            b[i+n]=e;

            op[i]=node(c,d,e);

        }

        printf("Case %d:\n",++k);

        sort(b+,b+n+m+);

        int num=unique(b+,b+n+m+)-b-;

        rep(i,,n)

        {

            a[i]=lower_bound(b+,b+num+,a[i])-b;

            insert(,num,root[i-],root[i],a[i]);

        }

        rep(i,,m)

        {

            int x=op[i].x,y=op[i].y,z=op[i].z;

            z=lower_bound(b+,b+num+,z)-b;

            printf("%d\n",query(,z,,num,root[x-],root[y]));

        }

    }

    //system("Pause");

    return ;

}