結果
| 問題 | 
                            No.2785 四乗足す四の末尾の0
                             | 
                    
| コンテスト | |
| ユーザー | 
                            👑  | 
                    
| 提出日時 | 2024-06-14 22:01:21 | 
| 言語 | C++17  (gcc 13.3.0 + boost 1.87.0)  | 
                    
| 結果 | 
                             
                                WA
                                 
                             
                            
                         | 
                    
| 実行時間 | - | 
| コード長 | 2,556 bytes | 
| コンパイル時間 | 4,315 ms | 
| コンパイル使用メモリ | 250,256 KB | 
| 最終ジャッジ日時 | 2025-02-21 21:59:50 | 
| 
                            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 | 
|---|---|
| other | AC * 7 WA * 13 | 
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" : " "); 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> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(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 solve(){
	long long n;
	cin >> n;
	if(n == 1 or n == -1){
		cout << "Yes\n";
		cout << "0\n";
		return 0;
	}
	if(n % 50 == 6 or n % 50 == 8 or n % 50 == 42 or n % 50 == 44){
		cout << "No\n";
		cout << "2\n";
		return 0;
	}
	if(n % 2 == 0 and n % 10 != 0){
		cout << "No\n";
		cout << "1\n";
		return 0;
	}
	{
		cout << "No\n";
		cout << "0\n";
		return 0;
	}
	
}
int main(){
	int t;
	cin >> t;
	rep(i, t) solve();
	return 0;
}