Problem Description

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.

 

Input

The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)

 

Output

For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.

 

Sample Input

 

Sample Output

 

#include <stdio.h>

#include <bits/stdc++.h>

using namespace std;

#define CLR(arr,val) memset(arr,val,sizeof(arr))

#define MID(x,y) ((x+y)>>1)

const int N=1e5+7;

struct seg

{

    int lson,rson;

    int cnt;

};

seg T[N*20];

int root[N],arr[N],tot;

vector<int>vec;

void init()

{

    CLR(root,0);

    tot=0;

    vec.clear();

}

void update(int &cur,const int &ori,const int &l,const int &r,const int &val)

{

    cur=++tot;

    T[cur]=T[ori];

    ++T[cur].cnt;

    if(l==r)

        return ;

    int mid=MID(l,r);

    if(val<=mid)

        update(T[cur].lson,T[ori].lson,l,mid,val);

    else

        update(T[cur].rson,T[ori].rson,mid+1,r,val);

}

int query(const int &S,const int &E,const int &l,const int &r,const int &x,const int &y)

{

    if(x<=l&&r<=y)//l<=T[k].l&&T[k].r<=r

        return T[E].cnt-T[S].cnt;//return T[k].cnt;

    int mid=MID(l,r);

    if(y<=mid)//r<=T[k].mid

        return query(T[S].lson,T[E].lson,l,mid,x,y);

    else if(x>mid)//l>T[k].mid

        return query(T[S].rson,T[E].rson,mid+1,r,x,y);

    else

        return query(T[S].lson,T[E].lson,l,mid, x,mid)+query(T[S].rson,T[E].rson,mid+1,r, mid+1,y);

}

int main(void)

{

    int T,n,m,i,l,r,h;

    scanf("%d",&T);

    for (int q=1; q<=T; ++q)

    {

        init();

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

        for (i=1; i<=n; ++i)

        {

            scanf("%d",&arr[i]);

            vec.push_back(arr[i]);

        }

        sort(vec.begin(),vec.end());

        vec.erase(unique(vec.begin(),vec.end()),vec.end());

        int R=vec.size();

        for (i=1; i<=n; ++i)

        {

            arr[i]=lower_bound(vec.begin(),vec.end(),arr[i])-vec.begin()+1;

            update(root[i],root[i-1],1,R,arr[i]);

        }

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

        for (i=0; i<m; ++i)

        {

            scanf("%d%d%d",&l,&r,&h);

            ++l;

            ++r;

            int indx=lower_bound(vec.begin(),vec.end(),h)-vec.begin()+1;

            //由于H可能过大因此只能找一个最接近的值indx来等效代替

            if(vec[indx-1]!=h)

                --indx;

            printf("%d\n",indx? query(root[l-1],root[r],1,R,1,indx) : 0);

        }

    }

    return 0;

}