题解

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1005;
const int MOD = 998244353;
#define endl '\n'

int h[N];
int f[N][40005];

int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &h[i]);
    const int OFFSET = 20005;
    ll ans = n;  // 单塔方案
    // f[i][d]: 以 i 结尾,公差 d 的等差子序列数(长度 ≥ 2)
    for(int i = 1; i <= n; i++){
        for(int j = 1; j < i; j++){
            int d = h[i] - h[j] + OFFSET;
            f[i][d] = (f[i][d] + f[j][d] + 1) % MOD;
        }
        for(int d = 0; d < 40005; d++){
            ans = (ans + f[i][d]) % MOD;
        }
    }
    printf("%lld\n", ans);
    return 0;
}