結果
問題 | No.2734 Addition and Multiplication in yukicoder (Hard) |
ユーザー |
👑 |
提出日時 | 2024-04-20 19:08:02 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 1,209 ms / 5,000 ms |
コード長 | 1,822 bytes |
コンパイル時間 | 4,458 ms |
コンパイル使用メモリ | 256,572 KB |
最終ジャッジ日時 | 2025-02-21 07:12:46 |
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 36 |
ソースコード
#include<bits/stdc++.h>#include<atcoder/all>#define rep(i,n) for(int i=0;i<n;i++)using namespace std;using namespace atcoder;typedef long long ll;typedef vector<int> vi;typedef vector<long long> vl;typedef vector<vector<int>> vvi;typedef vector<vector<long long>> vvl;typedef long double ld;typedef pair<int, int> P;ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); returnos;}template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : "");return os;}template<typename T> void chmin(T& a, T b){a = min(a, b);}template<typename T> void chmax(T& a, T b){a = max(a, b);}using mint = modint998244353;int main(){int n;cin >> n;vector<long long> a(n);cin >> a;sort(a.begin(), a.end(), [](long long left, long long right){string s_left = to_string(left);string s_right = to_string(right);string s_left_right = s_left + s_right;string s_right_left = s_right + s_left;return s_left_right < s_right_left;});mint ans = 0;rep(i, n){int d = to_string(a[i]).size();ans *= mint(10).pow(d);ans += a[i];}cout << ans << "\n";return 0;}