結果

問題 No.2734 Addition and Multiplication in yukicoder (Hard)
ユーザー 👑 binap
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
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 36
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#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" : " "); return
    os;}
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;
}
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