題目:
Pie
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7795 Accepted Submission(s): 2889
My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.
What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.
---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.
---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.
描述:給N個等高圓柱體的半徑,求把他們分給F+1個人的最大體積
題解:二分裸題,但是要注意精度,1e-6正好A,1e-8蜜汁WA
代碼:
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const double eps = 1e-6;
const int maxn = 10005;
const double pi = acos(-1.0);
int r[maxn];
int main()
{
//freopen("input.txt", "r", stdin);
int T;
scanf("%d", &T);
while (T--)
{
int N, F;
scanf("%d%d", &N, &F);
for (int i = 0; i < N; i++)
{
scanf("%d", &r[i]);
}
sort(r, r + N);
double l = 0.0;
double r1 = pi*r[N - 1] * r[N - 1];
while (r1 - l > eps)
{
double mid = (l + r1) / 2;
int temp = 0;
for (int i = 0; i < N; i++)
temp += floor(pi*r[i] * r[i] / mid);
if (temp >= F + 1)
l = mid;
else
r1 = mid;
}
printf("%.4lf\n", l);
}
return 0;
}