#665443 (C++17) No.1543 [Cherry 2nd Tune A] これがプログラミングなのね

• 問題: No.1543 [Cherry 2nd Tune A] これがプログラミングなのね
• ユーザー: Anumish
• 提出日時: 2021-06-11 21:22:08
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 2 ms / 2,000 ms
• コード長: 2,124 bytes
• コンパイル時間: 831 ms
• コンパイル使用メモリ: 94,712 KB
• 最終ジャッジ日時: 2025-01-22 05:24:22
• ジャッジサーバーID （参考情報）: judge2 / judge2

ファイルパターン	結果
other	AC * 35

ソースコード

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//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("Ofast,unroll-loops")
//#pragma GCC target("avx,avx2,fma")
#include <algorithm>
#include <array>
#include <cassert>
//#include <chrono>
#include <cmath>
//#include <cstring>
//#include <functional>
//#include <iomanip>
#include <iostream>
#include <map>
//#include <numeric>
//#include <queue>
//#include <random>
#include <set>
#include <vector>
using namespace std;
#define FAST  ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
//#include <ext/pb_ds/assoc_container.hpp> 
//#include <ext/pb_ds/tree_policy.hpp> 
//using namespace __gnu_pbds; 
#define int long long 
#define ll int
#define all(a) a.begin(),a.end()    
#define rev(a) a.rbegin(),a.rend()    
//typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set; //less_equal for multiset
#define ar array
#define pb push_back
#define fi(a,b) for(int i=a;i<(b);i++)
#define fj(a,b) for(int j=a;j<(b);j++)
#define fk(a,b) for(int k=a;k<(b);k++)
const double pi=acosl(-1);
void solve()
{
    int x,y;
    char c;
    cin>>x>>y>>c;
    if(c=='<')
    {
        if(x<y)
        {
            cout<<"YES\n";
        }
        else
        {
            cout<<"NO\n";
        }
    }
    else if(c=='=')
    {
        if(x==y)
        {
            cout<<"YES\n";
        }
        else
        {
            cout<<"NO\n";
        }
    }
    else
    {
        if(x>y)
        {
            cout<<"YES\n";
        }
        else{
            cout<<"NO\n";
        }
    }
    
    
}
signed main()
{
    FAST;
    int tt=1;
    //cin>>tt;    
    while(tt--)
    {
        solve();
    }    
}
//int dx[] = {+1,-1,+0,+0,-1,-1,+1,+1}; // Eight Directions
//int dy[] = {+0,+0,+1,-1,+1,-1,-1,+1}; // Eight Directions
//int dx[]= {-2,-2,-1,1,-1,1,2,2}; // Knight moves
//int dy[]= {1,-1,-2,-2,2,2,-1,1}; // Knight moves
// For (a^b)%mod, where b is large, replace b by b%(mod-1)..{because we solve this by introducing log ..do same for general cases}
// a+b = (a|b)+(a&b)
// a+b = 2*(a&b)+(a^b)
 
 
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