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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#501323#5152. Circular Caramel CookieGrunrayWA 0ms3816kbC++202.5kb2024-08-02 16:45:472024-08-02 16:45:47

Judging History

你现在查看的是最新测评结果

  • [2024-08-02 16:45:47]
  • 评测
  • 测评结果:WA
  • 用时:0ms
  • 内存:3816kb
  • [2024-08-02 16:45:47]
  • 提交

answer

#define _CRT_SECURE_NO_WARNINGS
#define itn int
#define PII pair<int, int>
#define PLI pair<long long, int>
#define fep(i, a, b) for(int i = (a); i >= (b); --i)
#define rep(i, a, b) for(int i = (a); i <= (b); ++i)
#include<bits/stdc++.h>
#include<unordered_map>
using ll = long long;
using ldou = long double;
using unll = unsigned long long;
using namespace std;

inline int read() {
	int x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch>'9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
	return x * f;
}

ll gcd(ll a, ll b) { // 最大公约数
	while (b ^= a ^= b ^= a %= b)
		;
	return a;
}
ll lcm(ll a, ll b) { // 最小公倍数
	return a / gcd(a, b) * b;
}
ll qmi(ll m, ll k, ll p) { // 快速幂
	//求 m^k mod p,时间复杂度 O(logk)。
	//m为底数,k为幂
	ll res = 1 % p, t = m;
	while (k) {
		if (k & 1) res = res * t % p;
		t = t * t % p;
		k >>= 1;
	}
	return res;
}
unll qmi(unll m, unll k, unll p) { //龟速乘
	ll res = 0, t = m;
	while (k) {
		if (k & 1) res = (res + t) % p;
		k >>= 1;
		t = (t << 1) % p;
	}
	return res;
}

////////////////////////////////////////////////////////////////////////////////

ll n, m, k;
const int N = 1e4 + 50;
const int M = 2e5 + 50;
const ll INF = 1e9;
const ll MODE = ll(998244353);
const double Pi = 3.1415926;
const double eps = 1e-6;
const int dx[4] = { 1,-1, 0, 0 };
const int dy[4] = { 0, 0, 1,-1 };
//priority_queue<int> p;//这是一个大根堆q
//priority_queue<int, vector<int>, greater<int> >q;//这是一个小根堆q
//priority_queue<ll, vector<ll>, greater<ll> >pq; // 小根

ll s;

bool check(double mid) {
	int cnt = 0;
	int ans = 0;
	for (int i = 1; i <= mid; i++) {
		cnt = sqrt(mid * mid - i * i);
		ans += cnt;
	}
	ans *= 4;
	return ans >= s;
}

double bser(double l, double r) {
	while (r - l > eps) {
		double mid = (l + r) / 2;
		if (check(mid)) 
			r = mid;
		else 
			l = mid;
		//cout << l << ' ' << r << '\n';
	}
	return l;
}

void solve() {

	cin >> s;

	if (s == 1) {
		cout << 1.0000000000 << '\n';
		return;

	}

	double l = 0, r = sqrt(s);
	
	cout << fixed << setprecision(10) << (double)bser(l, r) << '\n';







	

}

signed main() {
	std::ios::sync_with_stdio(false); std::cin.tie(0), std::cout.tie(0);
	/*freopen("out.txt", "r", stdin);
	freopen("wrt.txt", "w", stdout);*/
	int TTT = 1; //cin >> TTT;
	while (TTT--) {
		solve();
	}
	/*while (cin >> n >> m) {
		solve();
	}*/

	return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 0ms
memory: 3692kb

input:

11

output:

2.2360678352

result:

ok found '2.2360678', expected '2.2360680', error '0.0000001'

Test #2:

score: 0
Accepted
time: 0ms
memory: 3648kb

input:

59

output:

4.9999993583

result:

ok found '4.9999994', expected '5.0000000', error '0.0000001'

Test #3:

score: -100
Wrong Answer
time: 0ms
memory: 3816kb

input:

1

output:

1

result:

wrong answer 1st numbers differ - expected: '1.4142136', found: '1.0000000', error = '0.2928932'