10分：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 50010;
int n , q;
int c[N] , cnt[N];
ll ans , ans2[N] , res[N];
ll gcd (ll a , ll b) {
  return b ? gcd (b , a % b) : a;
}
array<int , 3> que[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL) , cout.tie(NULL);
  cin >> n >> q;
  // scanf ("%d%d" , &n , &q);
  for (int i = 1; i <= n; ++ i) {
    cin >> c[i];
  }
  for (int i = 0; i < q; ++ i) {
    int l , r;
    cin >> l >> r;
    // scanf ("%d%d" , &l , &r);
    que[i] = {l , r , i};
    ans2[i] = (ll)(r - l) * (r - l + 1) / 2;
  }
  int B = 500;
  sort (que , que + q , [&](array<int , 3> a , array <int  , 3> b) {
    if (a[0] / B != b[0] / B) return a[0] / B  < b[0] / B;
    return a[1] < b[1];
  });
  int l = 1 , r = 0;
  auto add = [&] (int x) {
    ans += cnt[c[x]];
    cnt[c[x]] ++;
  };
  auto del = [&] (int x) {
    cnt[c[x]] --;
    ans -= cnt[c[x]];
  };
  for (int i = 0; i < q; ++ i) {
    while (r < que[i][1]) {
      r++;
      add (r);
    }
    while (l > que[i][0]) {
      l --;
      add (l);
    }
    while (r > que[i][1]) {
      del (r);
      r--;
    }
    while (l < que[i][0]) {
      del (l);
      l ++;
    }
    res[que[i][2]] = ans;
  }
  for (int i = 0; i < q; ++ i) {
    ll d = gcd (res[i] , ans2[i]);
    if (que[i][1] == que[i][0]/*这里*/) {
      cout << "0/1" << endl;
      continue;
    }
    if (d == 0) d = 1;
    cout << res[i] / d << '/' << ans2[i] / d << endl;
    // printf("%lld/%lld\n", res[i] / d , ans2[i] / d);
  }
  return 0;
}

100

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 50010;
int n , q;
int c[N] , cnt[N];
ll ans , ans2[N] , res[N];
int tl[N] , tr[N];
ll gcd (ll a , ll b) {
  return b ? gcd (b , a % b) : a;
}
array<int , 3> que[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL) , cout.tie(NULL);
  cin >> n >> q;
  // scanf ("%d%d" , &n , &q);
  for (int i = 1; i <= n; ++ i) {
    cin >> c[i];
  }
  for (int i = 0; i < q; ++ i) {
    int l , r;
    cin >> l >> r;
    // scanf ("%d%d" , &l , &r);
    que[i] = {l , r , i};
    tl[i] = l , tr[i] = r;
    ans2[i] = (ll)(r - l) * (r - l + 1) / 2;
  }
  int B = 500;
  sort (que , que + q , [&](array<int , 3> a , array <int  , 3> b) {
    if (a[0] / B != b[0] / B) return a[0] / B  < b[0] / B;
    return a[1] < b[1];
  });
  int l = 1 , r = 0;
  auto add = [&] (int x) {
    ans += cnt[c[x]];
    cnt[c[x]] ++;
  };
  auto del = [&] (int x) {
    cnt[c[x]] --;
    ans -= cnt[c[x]];
  };
  for (int i = 0; i < q; ++ i) {
    while (r < que[i][1]) {
      r++;
      add (r);
    }
    while (l > que[i][0]) {
      l --;
      add (l);
    }
    while (r > que[i][1]) {
      del (r);
      r--;
    }
    while (l < que[i][0]) {
      del (l);
      l ++;
    }
    res[que[i][2]] = ans;
  }
  for (int i = 0; i < q; ++ i) {
    ll d = gcd (res[i] , ans2[i]);
    if (tl[i] == tr[i]/*这里*/) {
      cout << "0/1" << endl;
      continue;
    }
    if (d == 0) d = 1;
    cout << res[i] / d << '/' << ans2[i] / d << endl;
    // printf("%lld/%lld\n", res[i] / d , ans2[i] / d);
  }
  return 0;
}

两份代码就只有特判 $l,r$ 相等的地方不一样。