题目链接：HDU4417

【题意】给出N长度的序列，Q个询问(x,y,h)；求出区间[x,y]中<=h的有多少个。

【分析】其实就是求h在区间[x,y]的rank值，可以用主席树记录每个区间各个值出现的次数，离线+离散即可，统计的时候往右走加上左边的数量，就可以了。

【AC CODE】171ms

#include <cstdio>
#include <cstring>
#include <cctype>
#include <cmath>
#include <set>
//#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
typedef long long LL;
#define rep(i,a,n) for(int i = a; i < n; i++)
#define repe(i,a,n) for(int i = a; i <= n; i++)
#define per(i,n,a) for(int i = n; i >= a; i--)
#define clc(a,b) memset(a,b,sizeof(a))
const int INF = 0x3f3f3f3f, MAXN = 200000+10;
struct IN{
    int x,y,h;
}in[100000+10];
int sum[MAXN*20],lc[MAXN*20], rc[MAXN*20], cnt, n,q,a[MAXN/2],rt[MAXN/2],num[MAXN],sot[MAXN];
int tol;

void bulid(int &u, int x, int y)
{
    sum[cnt] = 0;u = cnt++;
    if(x == y) return;
    int m = (x+y)>>1;
    bulid(lc[u],x,m);
    bulid(rc[u],m+1,y);
}
void insert(int &u, int x, int y, int p)
{
    sum[cnt] = sum[u], lc[cnt] = lc[u], rc[cnt] = rc[u];
    u = cnt++;sum[u]++;
    if(x == y) return;
    int m = (x+y)>>1;
    if(p <= m) insert(lc[u],x,m,p);
    else insert(rc[u],m+1,y,p);
}
int mhash(int v){return lower_bound(sot+1,sot+1+tol,v)-sot;}
void init()
{
    cnt = 0;
    bulid(rt[0],0,n+q-1);
    sort(num,num+n+q);
    tol = 0;
    sot[++tol] = num[0];
    rep(i,1,n+q)
    {
        if(num[i] != num[i-1]) sot[++tol] = num[i];
    }
    repe(i,1,n)
    {
        int p = mhash(a[i]);
        rt[i] = rt[i-1];
        insert(rt[i],1,tol,p);
    }
}
int query(int ta, int tb, int x, int y,int p)
{
    if(x == y) return sum[tb]-sum[ta];
    int m = (x+y)>>1;
    if(p <= m) return query(lc[ta],lc[tb],x,m,p);
    return query(rc[ta], rc[tb],m+1,y,p)+sum[lc[tb]]-sum[lc[ta]];
}
int main()
{
#ifdef SHY
    freopen("d:\\1.txt", "r", stdin);
#endif
    int t,count = 0;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &n, &q);
        repe(i,1,n) scanf("%d", &a[i]),num[i-1] = a[i];
        rep(i,0,q)
        {
            int x,y,h;
            scanf("%d%d%d", &x, &y, &h);
            in[i] = {x+1,y+1,h};
            num[n+i] = h;
        }
        init();
        printf("Case %d:\n",++count);
        rep(i,0,q)
        {
            printf("%d\n", query(rt[in[i].x-1], rt[in[i].y],1,tol,mhash(in[i].h)));
        }
    }
    return 0;
}