結果

問題 No.2692 How Many Times Reached?
ユーザー 👑 binapbinap
提出日時 2024-03-22 21:27:06
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 2,001 bytes
コンパイル時間 4,539 ms
コンパイル使用メモリ 263,584 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-09-30 10:49:11
合計ジャッジ時間 5,027 ms
ジャッジサーバーID
(参考情報)
judge3 / judge1
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 * 4
other AC * 43
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);}
int main(){
int n;
cin >> n;
vector<string> s(n);
cin >> s;
int ans = 0;
rep(y, n){
int p = 0, q = 0;
rep(x, n){
if(s[y][x] == 'A') p++;
if(s[y][x] == '.') q++;
}
if(p == n - 1 and q == 1) ans++;
}
rep(x, n){
int p = 0, q = 0;
rep(y, n){
if(s[y][x] == 'A') p++;
if(s[y][x] == '.') q++;
}
if(p == n - 1 and q == 1) ans++;
}
{
int p = 0, q = 0;
rep(i, n){
if(s[i][i] == 'A') p++;
if(s[i][i] == '.') q++;
}
if(p == n - 1 and q == 1) ans++;
}
{
int p = 0, q = 0;
rep(i, n){
if(s[i][(n - 1) - i] == 'A') p++;
if(s[i][(n - 1) - i] == '.') q++;
}
if(p == n - 1 and q == 1) ans++;
}
cout << ans << "\n";
return 0;
}
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