結果

問題 No.2239 Friday
ユーザー omgflakaomgflaka
提出日時 2023-03-10 21:36:05
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 2,383 bytes
コンパイル時間 2,145 ms
コンパイル使用メモリ 200,640 KB
実行使用メモリ 4,372 KB
最終ジャッジ日時 2023-10-18 07:18:00
合計ジャッジ時間 2,979 ms
ジャッジサーバーID
（参考情報） 		judge15 / judge13
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 WA -
testcase_01 AC 2 ms
4,368 KB
testcase_02 WA -
testcase_03 WA -
testcase_04 WA -
testcase_05 AC 2 ms
4,368 KB
testcase_06 WA -
testcase_07 WA -
testcase_08 AC 2 ms
4,368 KB
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 WA -
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 # 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define vi vector<int> 
#define vb vector<bool> 
#define pub push_back
#define emb emplace_back
#define rep(i,a,b) for(long long i = a; i < b; i++)
#define rrep(i,a,b) for(long long i = a; i >= b; i--)
#define iterm(key,val,ok) for(auto const& [key, val]:ok)
#define all(x) x.begin(),x.end()
#define py cout<<"YES"<<endl
#define pn cout<<"NO"<<endl
#define pa cout<<ans<<endl
#define pe cout<<endl
#define mp make_pair
#define f first 
#define se second 
using pii=pair<ll,ll>;
template<typename T>             istream& operator>>(istream& is, vector<T>& v) { for (auto& i : v) is >> i;        return is; }
template<typename T>             ostream& operator<<(ostream& os, const vector<T>& v) { for (auto& i : v) os << i << " "; return os; }
template<typename T>             istream& operator>>(istream& is, pair<T, T>& v) { is >> v.first >> v.second;        return is; }
template<typename T>             ostream& operator<<(ostream& os, const pair<T, T>& v) { os << v.first << " " << v.second; return os; }
 
 
 
/*
problems with mod or small constraints are DP (2d,3d) or combo
ascii: 1=49 a=97 A=65
check edge cases and constraints CHECK FOR JUST ONE ELEMENT OR SAME NUMBERS ETC
always think about reversing a process
see if theres any dumb tricks if youre not getting it: like only having to check small values or sth
(a-b)modm = (a%m+(m-b%m))%m
mark visited after adding to queue
think about transforming representations of graphs, if some graph is too big to be represented, think about bipartite
representation, if its too small maybe you can expand according to problem to make it easier
you can also bfs on pairs of nodes, different representations of nodes etc etc
lower bound can give beyond end also (check segfaults etc if using)
many smol vectors are slowwww, use std:array instead
to search for and remove particular things fastly just use multiset, like u can use prefix multiset and find closest
value of something also like so many applications in just logn time
*/
 

void solve(){
    ll a,b;cin>>a>>b;
    if(a>b){
        swap(a,b);
    }
    if(b>=(2*a)){
        cout<<(b-2*a)<<endl;
    }
    else{
        cout<<b-a<<endl;
    }
}
 
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
}
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