#464. 数数

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whiteawll 2022-08-16 14:50:43 博主文章分类：练习题目 ©著作权

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#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
#define int long long
namespace ls {
    const int maxnum = 5e5 + 5;
    int C[maxnum], L[maxnum], A[maxnum];
    //把 长度为n的a数组的内容离散化到 f 中
    void ls(int *f, int *a, int n) {
        for (int i = 0; i < n; i++)A[i] = a[i + 1];
        memcpy(C, A, sizeof(A));
        sort(C, C + n);
        int l = unique(C, C + n) - C;
        for (int i = 0; i < n; ++i)
            L[i] = lower_bound(C, C + l, A[i]) - C + 1;
        for (int i = 0; i < n; i++)f[i + 1] = L[i];
    }
}
const int M = 5e5 + 10;

int n, q;
int a[M], f[M], ans[M], b[M];
struct Node {
    int k, x, id;
};
vector<Node> g[M];
int L[M], R[M], H[M];

int lowbit (int x) {
    return x & -x;
}
void update (int x, int y) {
    for (;x <= n + q; x += lowbit(x))
        b[x] += y;
}
int query (int x) {
    int ans = 0;
    for (;x; x -= lowbit(x))
        ans += b[x];
    return ans;
}
void solve() {
    memset(ans, 0, sizeof ans);
    memset(b, 0, sizeof b);
    memset(a, 0, sizeof a);
    memset(f, 0, sizeof f);
    n = read(), q = read();
    for (int i =1; i<= n;i ++)
        a[i] = read();
    
    rep(i, 1, q)
    L[i] = read(),R[i] = read(), a[n + i] = read();
    
    rep(i, 1, q +n)
        g[i].clear();
    
    ls::ls(f, a, n + q);
    rep(i, 1, n) 
        a[i] = f[i];
    rep(i, 1, q) 
        H[i] = f[n + i];
    
    rep(i, 1, q) {
        g[L[i] - 1].pb({-1, H[i], i});
        g[R[i]].pb({1, H[i], i});
    }
    rep(i, 1, n) {
        update(a[i], 1);
        for (auto v : g[i])
            ans[v.id] += v.k * query(v.x);
    }
    rep(i, 1, q)
        printf("%lld ", ans[i]);
    puts(""); 
}
signed main () {
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}