結果

問題 No.2244 Integer Complete
ユーザー Jeroen Op de BeekJeroen Op de Beek
提出日時 2023-03-10 23:18:55
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 2,264 bytes
コンパイル時間 2,158 ms
コンパイル使用メモリ 198,476 KB
最終ジャッジ日時 2025-02-11 09:15:12
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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ファイルパターン 結果
other AC * 42 WA * 18
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;
#define all(x) begin(x),end(x)
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = " "; return os; }
#define debug(a) cerr << "(" << #a << ": " << a << ")\n";
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pi;
const int mxN = 1e5+1, oo = 1e9;
int main() {
    // ok, what can I not do?
    // some integer such that
    // prime you cannot make at least...
    // 1e4 + something to be prime, it's impossible to make
    // so can just brute force check 1e4 candidates.
    // for each candidate, factorize
    int n,m; cin >> n >> m;
    vi a(n),b(m);
    for(auto& i : a) cin >> i;
    for(auto& i : b) cin >> i;
    // can make a lot more numbers than I thought
    // some prime such that it's not within those and not within those?
    // actually a bit like convolution.
    // but multiplicative.
    // answer not bigger than 1e9+7.
    // if you just constantly do the least thing not being able
    // ok, a prime number has to be made by 1 p, or p 1.
    // so if there's some gap, than immediately smaller than the gap you win
    // everything up to that can be made
    vi cnt(max(*max_element(all(a)), *max_element(all(b))) + 100);
    for(int i : a) cnt[i]++;
    for(int j : b) cnt[j]++;
    int i=1;
    if(cnt[1]==2) {
        while(cnt[i]) ++i;
    }
    // now restrict to here
    // need to test ~1e4 candidates
    for(ll c=i*i;;c++) {
        bool gd=0;
        auto check = [&](ll i, ll j) {
            i = sqrtl(i+0.5), j =sqrtl(j+0.5);
            cnt[i]--;
            cnt[j]--;
            if(cnt[i]>=0 and cnt[j]>=0) gd=1;
            cnt[i]++,cnt[j]++;
        };
        for(ll j=1;j*j<=c;++j) {
            if(c%j==0) {
                check(j,c/j);
                check(c/j,j);
            }
        }
        if(!gd) {
            cout << c << '\n';
            break;
        }
    }
}
0