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
1
10 10
0 5 2 7 5 4 3 8 7 7
2 8 6
3 5 0
1 3 1
1 9 4
0 1 0
3 5 5
5 5 1
4 6 3
1 5 7
5 7 3

Sample Output
Case 1:
4
0
0
3
1
2
0
1
5
1

解法1： 线段树/BIT离线
1，先把所有位置的高度都存下来，然后排序，注意保存下标；2，把所有询问存下来，然后按照询问的高度进行排序，同注意保存下标；3，对于排序后的每次询问的处理：由于每个位置的高度都已经存了下来并且进行了排序，所以可以按照顺序将每个点插入到线段树的对应位置(保存的下标)，并更新 线段树，直到要插入的位置的高度大于这次询问的高度H；最后处理区间查询，由于刚才已经把小于等于该次查询高度的位置都已经插入到了线段树中，所以询问的 结果就是查询区间中被更新过的叶子节点的个数，也就是区间求和问题。当然换成BIT也完全一样

//156ms 4.9Mb
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
struct node{
    int h, pos;
    node(){}
    node(int h, int pos) : h(h), pos(pos){}
    bool operator < (const node &rhs) const{
        return h < rhs.h;
    }
}a[maxn];

struct Q{
    int l, r, h, id;
    Q(){}
    Q(int l, int r, int h, int id) : l(l), r(r), h(h), id(id){}
    bool operator < (const Q &rhs) const{
        return h < rhs.h;
    }
}q[maxn];

int ans[maxn];
namespace segmenttree{
    int sum[maxn*4];
    inline void init(){memset(sum, 0, sizeof(sum));}
    inline void pushup(int o){sum[o] = sum[o*2] + sum[o*2+1];}
    inline void update(int pos, int l, int r, int o){
        if(l == r){
            sum[o]++;
            return ;
        }
        int m = (l + r) / 2;
        if(pos <= m) update(pos, l, m, o*2);
        else update(pos, m + 1, r, o*2+1);
        pushup(o);
    }
    inline int query(int L, int R, int l, int r, int o){
        if(L <= l && r <= R) return sum[o];
        int m = (l + r) / 2;
        if(R <= m) return query(L, R, l, m, o*2);
        else if(L > m) return query(L, R, m + 1, r, o*2+1);
        else return query(L, m, l, m, o*2) + query(m + 1, R, m + 1, r, o*2+1);
    }
}
using namespace segmenttree;
int main(){
    int n, m, T, ks = 0;
    scanf("%d", &T);
    while(T--){
        printf("Case %d:\n", ++ks);
        init();
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) scanf("%d", &a[i].h), a[i].pos = i;
        for(int i = 1; i <= m; i++){
            scanf("%d%d%d", &q[i].l, &q[i].r, &q[i].h); q[i].id = i;
        }
        sort(a + 1, a + n + 1);
        sort(q + 1, q + m + 1);
        int i, j;
        for(i = j = 1; i <= m; i++){
            int id = q[i].id, l = q[i].l, r = q[i].r;
            while(a[j].h <= q[i].h && j <= n){
                update(a[j++].pos, 1, n, 1);
            }
            ans[id] = query(l + 1, r + 1, 1, n, 1);
        }
        for(int k = 1; k <= m; k++){
            printf("%d\n", ans[k]);
        }
    }
    return 0;
}

解法2：在线主席树做法，我们知道主席数可以方便的维护区间第k大，那么我们可以在此基础上统计第i小的的数小于k的个数。

//156ms 11.1MB
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
const int maxm = 30*maxn;
int n, m, q, tot, a[maxn], b[maxn];
int getid(int x){return lower_bound(b + 1, b + m + 1, x) - b; }//下标从1开始
namespace chairmantree{
    int T[maxm], lson[maxm], rson[maxm], c[maxm];
    int build(int l, int r){
        int root = tot++;
        c[root] = 0;
        if(l != r){
            int mid = (l + r) / 2;
            lson[root] = build(l, mid);
            rson[root] = build(mid + 1, r);
        }
        return root;
    }
    int update(int root, int pos, int val){
        int newroot = tot++, tmp = newroot;
        c[newroot] = c[root] + val;
        int l = 1, r = m;
        while(l < r){
            int mid = (l + r) / 2;
            if(pos <= mid){
                lson[newroot] = tot++, rson[newroot] = rson[root];
                newroot = lson[newroot], root = lson[root], r = mid;
            }
            else{
                rson[newroot] = tot++, lson[newroot] = lson[root];
                newroot = rson[newroot], root = rson[root], l = mid + 1;
            }
            c[newroot] = c[root] + val;
        }
        return tmp;
    }
    int query(int L, int R, int k){
        int res = 0;
        int l = 1, r = m;
        while(l < r){
            int mid = (l + r) / 2;
            if(k <= mid){
                r = mid;
                L = lson[L];
                R = lson[R];
            }
            else{
                l = mid + 1;
                res += c[lson[R]] - c[lson[L]];
                L = rson[L];
                R = rson[R];
            }
        }
        return res + (k < l ? 0 : c[R] - c[L]);
    }
}
using namespace chairmantree;

int main(){
    int TT, ks = 0;
    scanf("%d", &TT);
    while(TT--){
        printf("Case %d:\n", ++ks);
        scanf("%d%d", &n, &q);
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(int i = 1; i <= n; i++) b[i] = a[i];
        sort(b + 1, b + n + 1);
        m = unique(b + 1, b + n + 1) - b - 1;
        T[0] = build(1, m);
        for(int i = 1; i <= n; i++) T[i] = update(T[i-1], getid(a[i]), 1);
        while(q--){
            int l, r, h;
            scanf("%d%d%d", &l, &r, &h);
            h = upper_bound(b+1, b+m+1, h) - b - 1;//有重复元素,所以要找upper_bound
            printf("%d\n", query(T[l], T[r+1], h));
        }
    }
    return 0;
}