結果
問題 | No.1459 スマホを落としたいだけなのに |
ユーザー |
|
提出日時 | 2021-03-31 21:35:42 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 2,513 bytes |
コンパイル時間 | 879 ms |
コンパイル使用メモリ | 95,272 KB |
実行使用メモリ | 6,824 KB |
最終ジャッジ日時 | 2024-12-15 15:57:02 |
合計ジャッジ時間 | 1,701 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 6 |
other | AC * 30 |
ソースコード
#include<iostream>#include<string>#include<algorithm>#include<cmath>#include<ctime>#include<map>#include<vector>#include<math.h>#include<stdio.h>#include<stack>#include<queue>#include<tuple>#include<cassert>#include<set>#include<bitset>#include<functional>#include <fstream>//#include<bits/stdc++.h>#pragma GCC target("avx2")#pragma GCC optimize("O3")#pragma GCC optimize("unroll-loops")#define rep(i, x) for(ll i = 0; i < x; i++)#define rep2(i, x) for(ll i = 1; i <= x; i++)#define all(a) (a).begin(),(a).end()#define puts(x) cout << (x) << "\n"using ll = long long;using ld = long double;using namespace std;const ll INF = 1000000000000000000;const int intINF = 1000000000;const ll mod = 1000000007;const ll MOD = 998244353;const ld pi = 3.141592653589793238;//const ld EPS = 1e-9;bool isprime(int p) {if (p == 1) return false;for (int i = 2; i < p; i++) {if (p % i == 0) return false;}return true;}ll gcd(ll a, ll b) {if (a < b)swap(a, b);if (a % b == 0)return b;return gcd(b, a % b);}// 返り値: a と b の最大公約数// ax + by = gcd(a, b) を満たす (x, y) が格納される//main関数内に extGCD(a, b, x, y); でx, yに解が格納ll extGCD(ll a, ll b, ll& x, ll& y) {if (b == 0) {x = 1;y = 0;return a;}ll d = extGCD(b, a % b, y, x);y -= a / b * x;return d;}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}ll keta(ll n) {ll res = 0;while (n >= 1) {res += n % 10; n /= 10;}return res;}ll modpow(ll x, ll y) {ll res = 1;while (y) {if (y % 2) { res *= x; res %= mod; }x = x * x % mod; y /= 2;}return res;}ll nCk(ll n, ll k) {ll a = 1, b = 1;for (int h = n - k + 1; h <= n; h++) { a *= h; a %= mod; }for (int h = 1; h <= k; h++) { b *= h; b %= mod; }return a * modpow(b, mod - 2) % mod;}//printf("%.10f\n", n);typedef pair <ll, ll> P;typedef pair <ld, ll> pp;ll dx[4] = { 1, 0, -1, 0 }, dy[4] = { 0, 1, 0, -1 };struct edge { ll to, cost; };struct status {ll ima;ll cost;bool operator<(const status& rhs) const { return cost < rhs.cost; };bool operator>(const status& rhs) const { return cost > rhs.cost; };};signed main() {ios::sync_with_stdio(false);std::cin.tie(nullptr);//cout << fixed << setprecision(15);//inputll n; cin >> n;ll ok = -1, ng = n + 1, mid = (ok + ng) / 2;while (ng - ok > 1) {ll cnt = mid * (mid + 1) / 2;if (cnt >= n) { ng = mid; }else { ok = mid; }mid = (ok + ng) / 2;}cout << ng << endl;return 0;}