結果

問題 No.1991 Secret Garden
ユーザー xyz2606
提出日時 2022-06-24 22:57:40
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,228 bytes
コンパイル時間 2,132 ms
コンパイル使用メモリ 146,608 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-11-08 18:41:27
合計ジャッジ時間 3,120 ms
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
sample AC * 1
other AC * 18 WA * 4
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 #
プレゼンテーションモードにする

#include<vector>
#include<set>
#include<map>
#include<queue>
#include<string>
#include<algorithm>
#include<iostream>
#include<bitset>
#include<functional>
#include<chrono>
#include<numeric>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cassert>
#include<cmath>
#include<iomanip>
#include<random>
#include<ctime>
#include<complex>
#include<type_traits>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef long long LL;
typedef double D;
#define all(v) (v).begin(), (v).end()
mt19937 gene(chrono::system_clock::now().time_since_epoch().count());
typedef complex<double> Complex;
#define fi first
#define se second
#define ins insert
#define pb push_back
inline char GET_CHAR(){
const int maxn = 131072;
static char buf[maxn],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,maxn,stdin),p1==p2)?EOF:*p1++;
}
inline int getInt() {
int res(0);
char c = getchar();
while(c < '0') c = getchar();
while(c >= '0') {
res = res * 10 + (c - '0');
c = getchar();
}
return res;
}
inline LL fastpo(LL x, LL n, LL mod) {
LL res(1);
while(n) {
if(n & 1) {
res = res * (LL)x % mod;
}
x = x * (LL) x % mod;
n /= 2;
}
return res;
}
template<LL mod> struct Num {
LL a;
Num operator + (const Num & b) { return Num{(a + b.a) % mod}; }
Num operator - (const Num & b) { return Num{(a - b.a + mod) % mod}; }
Num operator * (const Num & b) { return Num{a * b.a % mod}; }
Num operator / (const Num & b) { return Num{a * fastpo(b.a, mod - 2, mod) % mod}; }
void operator += (const Num & b) {if((a += b.a) >= mod) a -= mod;}
void operator -= (const Num & b) {if((a -= b.a) < 0) a += mod;}
void operator *= (const Num & b) { a = a * b.a % mod; }
void operator /= (const Num & b) { a = a * fastpo(b.a, mod - 2, mod) % mod; }
void operator = (const Num & b) { a = b.a; }
void operator = (const LL & b) { a = b; }
};
template<LL mod> ostream & operator << (ostream & os, const Num<mod> & a) {
os << a.a;
return os;
}
LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a; }
inline string itoa(LL x, int width = 0) {
string res;
if(x == 0) res.push_back('0');
while(x) {
res.push_back('0' + x % 10);
x /= 10;
}
while((int)res.size() < width) res.push_back('0');
reverse(res.begin(), res.end());
return res;
}
const int _B = 131072;
char buf[_B];
int _bl = 0;
inline void flush() {
fwrite(buf, 1, _bl, stdout);
_bl = 0;
}
__inline void _putchar(char c) {
if(_bl == _B) flush();
buf[_bl++] = c;
}
inline void print(LL x, char c) {
static char tmp[20];
int l = 0;
if(!x) tmp[l++] = '0';
else {
while(x) {
tmp[l++] = x % 10 + '0';
x /= 10;
}
}
for(int i = l - 1; i >= 0; i--) _putchar(tmp[i]);
_putchar(c);
}
typedef double C;
struct P {
C x, y;
void scan() {
double _x, _y;
scanf("%lf%lf", &_x, &_y);
x = _x; y = _y;
}
void print() {
cout << '(' << x << ' ' << y << ')' << endl;
}
P operator + (const P & b) const { return P{x + b.x, y + b.y}; }
P operator - (const P & b) const { return P{x - b.x, y - b.y}; }
C operator * (const P & b) const { return x * b.y - y * b.x; }
C operator % (const P & b) const { return x * b.x + y * b.y; }
};
P operator * (const C & x, const P & b) { return P{x * b.x, x * b.y}; }
const int N = 300033;
const int LOG = 20;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
int n, m;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int rela[N];
int getr(int x) {
int p = x;
while(rela[p] != p) p = rela[p];
int p1 = p; p = x;
while(rela[p] != p) {
int p2 = rela[p];
rela[p] = p1;
p = p2;
}
return p1;
}
int main() {
LL n, i;
cin >> n;
for(i = 1; i <= n; i++) {
if(i == 1) {
if(n == 1) break;
else continue;
}
vector<LL> vec;
for(LL j = 1; j <= i; j++) vec.pb(n - j + 1);
LL cur = n - i;
for(int j = 0; j + 1 < (int)vec.size(); j++) {
LL mx = vec[j] - cur + vec[j + 1] - cur - 1;
cur -= mx;
if(cur <= 0) break;
}
if(cur <= 0) break;
}
cout << n - i << endl;
}
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