The 2025 ICPC Asia East Continent Online Contest (II)——D. Arcane Behemoths · znzryb
## 思路讲解
这个就是我们这个数在 rank=2 到 rank=i 的总和。
```cpp
FOR(i, 2, N) {
ll lans = 0;
lans = binpow(2, N - i) * (binpow(3, i - 1) - 1);
lans %= mod;
lans *= inv2;
lans %= mod;
lans *= A[i];
lans %= mod;
ans += lans;
ans %= mod;
}
```
rank=1 的时候的贡献。
```cpp
FOR(i, 1, N) {
ll lans = binpow(2, N - i) * A[i];
lans %= mod;
ans += lans;
ans %= mod;
}
```
## AC代码
[https://qoj.ac/submission/1345581](https://qoj.ac/submission/1345581)
::: 源代码
```cpp
// by Gospel_rock
#include <bits/stdc++.h>
#define all(vec) vec.begin(), vec.end()
#define PB push_back
#define PF push_front
#define EB emplace_back
#define EF emplace_front
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define lson(o) (o << 1)
#define rson(o) (o << 1 | 1)
#define SZ(a) ((long long)a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var << ":" << var << "\n";
#define cend cerr << "\n-----------\n"
#define fsp(x) fixed << setprecision(x)
#define minn(vec) *min_element(vec.begin(), vec.end())
#define maxx(vec) *max_element(vec.begin(), vec.end())
using namespace std;
#ifdef LOCAL
template <typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组,但是格式不是很好
template <typename T> void _print(vector<T> v) {
cerr << "[ ";
for (T i : v)
_print(i), cerr << " ";
cerr << "]";
}
template <typename T> void _print(set<T> s) {
cerr << "{ ";
for (T i : s)
_print(i), cerr << " ";
cerr << "}";
}
template <typename T, typename U> void _print(pair<T, U> p) {
cerr << "{";
_print(p.first);
cerr << ", ";
_print(p.second);
cerr << "}";
}
template <typename T, typename U> void _print(map<T, U> m) {
cerr << "{ ";
for (auto &p : m)
_print(p), cerr << " ";
cerr << "}";
}
// 打印两层数组,格式好一些
template <typename T> void _print(vector<vector<T>> v) {
cerr << "{\n";
for (auto i : v)
_print(i), cerr << "\n";
cerr << " }";
}
#endif
using ll = long long;
using ull = unsigned long long;
using ld = double;
using LD = long double;
using i128 = __int128;
using pdd = pair<ld, ld>;
using plb = pair<ll, bool>;
using pll = pair<ll, ll>;
using vll = vector<ll>;
using vb = vector<bool>;
using tll = tuple<ll, ll, ll>;
using vii = vector<int>;
using vpll = vector<pll>;
using vtll = vector<tll>;
using vdb = vector<ld>;
using vc = vector<char>;
using arr3 = array<ll, 3>;
using arr2 = array<ll, 2>;
using CD = complex<double>;
static constexpr ll MAXN = (ll)1e6 + 10,
INF = (ll)1e17 + 3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod = 998244353;
static constexpr double eps = 1e-8;
const double pi = acos(-1.0);
ll N, M, Q, X, K, T, lT;
inline void __();
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> lT;
while (lT--)
__();
return 0;
}
/*
*/
ll binpow(ll a, ll k) {
a %= mod;
ll res = 1;
if (k < 0)
return 1;
while (k) {
if (k & 1) {
res *= a;
res %= mod;
}
k >>= 1;
a *= a;
a %= mod;
}
return res;
}
inline void __() {
cin >> N;
vll A(N + 5);
FOR(i, 1, N) { cin >> A[i]; }
sort(A.begin() + 1, A.begin() + N + 1);
ll ans = 0;
ll inv2 = binpow(2, mod - 2);
FOR(i, 2, N) {
ll lans = 0;
lans = binpow(2, N - i) * (binpow(3, i - 1) - 1);
lans %= mod;
lans *= inv2;
lans %= mod;
lans *= A[i];
lans %= mod;
ans += lans;
ans %= mod;
}
// #ifdef LOCAL
// cerr << "OK/n";
// #endif
FOR(i, 1, N) {
ll lans = binpow(2, N - i) * A[i];
lans %= mod;
ans += lans;
ans %= mod;
}
ans %= mod;
if (ans < 0)
ans += mod;
cout << ans << "\n";
}
/*
AC
https://qoj.ac/submission/1345581
*/
```
:::
## 心路历程(WA,TLE,MLE……)