QOJ.ac
QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#293502 | #55. 欧几里得距离之和 | iee | 0 | 0ms | 0kb | C++11 | 688b | 2023-12-29 11:40:37 | 2023-12-29 11:40:38 |
answer
#include <bits/stdc++.h>
using namespace std;
using ld = long double;
constexpr int T = 100000;
constexpr ld pi = acos(-1);
int main() {
cin.tie(0)->sync_with_stdio(0);
int n;
cin >> n;
vector<int> x(n), y(n);
for (int i = 0; i < n; i++) {
cin >> x[i] >> y[i];
}
vector<ld> z(n);
ld sum = 0;
for (int i = 0; i < T; i++) {
auto t = polar<ld>(1, 2 * pi / T * i);
ld c = t.real(), s = t.imag();
for (int j = 0; j < n; j++) {
z[j] = c * x[j] + s * y[j];
}
sort(z.begin(), z.end());
for (int j = 1; j < n; j++) {
sum += (z[j] - z[j - 1]) * j * (n - j);
}
}
cout << fixed << setprecision(10) << sum / T * pi / 2 << "\n";
return 0;
}
详细
Subtask #1:
score: 0
Time Limit Exceeded
Test #1:
score: 0
Time Limit Exceeded
input:
3000 -802420 -321989 227507 956314 -460698 -819834 -479809 -341770 191520 109304 712327 -189558 -578326 -41090 282566 982266 -859119 686756 209058 -23298 -884994 -349898 -11358 182915 -507706 -81622 745434 575941 -374809 139274 810223 367608 960234 -197223 439081 573568 -275182 999306 -583036 -61808...
output:
result:
Subtask #2:
score: 0
Skipped
Dependency #1:
0%
Subtask #3:
score: 0
Skipped
Dependency #1:
0%
Subtask #4:
score: 0
Skipped
Dependency #1:
0%
Subtask #5:
score: 0
Skipped
Dependency #1:
0%