結果
問題 | No.2828 Remainder Game |
ユーザー |
|
提出日時 | 2024-08-04 02:59:48 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
RE
|
実行時間 | - |
コード長 | 2,550 bytes |
コンパイル時間 | 2,160 ms |
コンパイル使用メモリ | 202,248 KB |
最終ジャッジ日時 | 2025-02-23 20:58:14 |
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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ファイルパターン | 結果 |
---|---|
other | AC * 7 WA * 6 RE * 7 |
ソースコード
#include <bits/stdc++.h>using namespace std;#define rep(i, n) for(int i=0; i<n; i++)#define debug 0#define YES cout << "Yes" << endl;#define NO cout << "No" << endl;using ll = long long;using ld = long double;const int mod = 998244353;const int MOD = 1000000007;const double pi = atan2(0, -1);const int inf = 1 << 31 - 1;const ll INF = 1LL << 63 - 1;#include <time.h>#include <chrono>//vectorの中身を空白区切りで出力template<typename T>void printv(vector<T> v) {for (int i = 0; i < v.size(); i++) {cout << v[i];if (i < v.size() - 1) {cout << " ";}}cout << endl;}//vectorの中身を改行区切りで出力template<typename T>void print1(vector<T> v) {for (auto x : v) {cout << x << endl;}}//二次元配列を出力template<typename T>void printvv(vector<vector<T>> vv) {for (vector<T> v : vv) {printv(v);}}//vectorを降順にソートtemplate<typename T>void rsort(vector<T>& v) {sort(v.begin(), v.end());reverse(v.begin(), v.end());}//昇順priority_queueを召喚template<typename T>struct rpriority_queue {priority_queue<T, vector<T>, greater<T>> pq;void push(T x) {pq.push(x);}void pop() {pq.pop();}T top() {return pq.top();}size_t size() {return pq.size();}bool empty() {return pq.empty();}};//mod mod下で逆元を算出する//高速a^n計算(mod ver.)ll power(ll a, ll n) {if (n == 0) {return 1;}else if (n % 2 == 0) {ll x = power(a, n / 2);x *= x;x %= mod;return x;}else {ll x = power(a, n - 1);x *= a;x %= mod;return x;}}//フェルマーの小定理を利用ll modinv(ll p) {return power(p, mod - 2) % mod;}int main() {int N;cin >> N;queue<vector<int>> q;vector<int> ans;q.push({ 0,N-1,5 });int cnt = 0;while (ans.size() < 5 && cnt < 29) {cnt++;int mn = q.front()[0], mx = q.front()[1], num = q.front()[2];q.pop();int k = -1;int K = (mx - mn + 1)/2;cout << N << " " << K << endl;rep(i, K) {cout << mn + i << " ";k = max(k, mn + i);}cout << endl;int rep;cin >> rep;if (rep > 0) {if (k == mn) {rep(i, rep) {ans.push_back(mn);}}else {q.push({ mn, k, rep });}}if (rep < num) {if (k + 1 == mx) {rep(i, num - rep) {ans.push_back(mx);}}else {q.push({ k + 1, mx, num - rep });}}}int sum = 0;rep(i, 5) {sum += ans[i];if (ans[i] == 0) {sum += N;}}cout << 0 << " " << 1 << endl << sum << endl;}