算法

sub1

枚举$3$条木棍编号，暴力判断是否等腰，时间复杂度$O(n^3)$

sub2

枚举腰和底的编号，记录这个腰有多少个，注意特判底和腰长度相同的情况，时间复杂度$O(n^2)$

sub3

直接输出$C_n^3$

sub4

枚举腰长$d$，不难发现底的长度为$[1,2d-1]$，记录前缀和即可，同样要特判底和腰长度相同的情况。时
间复杂度$O(n)$

C++ 代码

#include <iostream>
#define int long long

using namespace std;

const int N = 2e5 + 10, MOD = 998244353;
const int inv2 = (MOD + 1) / 2, inv3 = (MOD + 1) / 3, inv6 = inv2 * inv3 % MOD;

int a[N], tax[N], sum[N];
int mx, ans;

signed main() {
    int n;
    cin >> n;

    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        tax[a[i]]++;
        mx = max(mx, a[i]);
    }

    for (int i = 1; i <= mx; ++i) sum[i] = (sum[i - 1] + tax[i]) % MOD;

    for (int i = 1; i <= mx; ++i) {
        int x = tax[i];
        if (x < 2) continue;
        ans = (ans + x * (x - 1) / 2 % MOD * (sum[min(i * 2 - 1, mx)] - x) % MOD 
              + x * (x - 1) % MOD * (x - 2) % MOD * inv6 % MOD) % MOD;
    }

    cout << ans << '\n';

    return 0;
}