//    int len = sqrt(1ll * n * n / m);
//    int len = n / (sqrt(m));
    int len = sqrt(n);

原题为AT_abc293_g，此题中在设置块长时如果使用上述被注释掉的两种写法就会有三个点 $RE$，而直接开 $\sqrt n$ 的块就不会有问题

#include <bits/stdc++.h>

using namespace std;
#define ll long long

const ll maxn = 2e5 + 5;
int a[maxn],  L[maxn], R[maxn], p[maxn];
ll res, cnt[maxn], ans[maxn], t[maxn];
struct node {
    int l, r, id;
} q[maxn], qt[maxn];

bool cmp(node x, node y) {
    if (p[x.l] != p[y.l]) {
        return x.l < y.l;
    } else if (p[x.l] & 1){
        return x.r > y.r;
    } else {
        return x.r < y.r;
    }
}

void solve(int x, ll v) {
    res -= t[cnt[x]];
    cnt[x] += v;
    res += t[cnt[x]];
}

void best_coder() {
    for (ll i = 3; i <= maxn - 5; ++i) {
        t[i] = t[i - 1] + (i - 2) * (i - 1) / 2;
    }

    int n, m;
    cin >> n >> m;

//    int len = sqrt(1ll * n * n / m);
//    int len = n / (sqrt(m));
    int len = sqrt(n);
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        p[i] = (i - 1) / len + 1;
    }
    int num = n / len + (n % len != 0);
    for (int i = 1; i <= num; ++i) {
        L[i] = R[i - 1] + 1;
        R[i] = i * len;
    }
    R[num] = n;

    int l = 0, r = 0;
    for (int i = 1; i <= m; ++i) {
        cin >> q[i].l >> q[i].r;
        q[i].id = i;
    }
    sort(q + 1, q + 1 + m, cmp);
    for (int i = 1; i <= m; ++i) {
        while (l < q[i].l) {
            solve(a[l++], -1);
        }
        while (l > q[i].l) {
            solve(a[--l], 1);
        }
        while (r < q[i].r) {
            solve(a[++r], 1);
        }
        while (r > q[i].r) {
            solve(a[r--], -1);
        }
        ans[q[i].id] = res;
    }
    for (int i = 1; i <= m; ++i) {
        cout << ans[i] << '\n';
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    best_coder();

    return 0;
}

附上完整代码