[HDU - 5954 ] Do not pour out（ 几何，二分法，积分）
题意：
你有一个圆柱杯。底部直径为2个单位，高度为2个单位。
杯中的液位高度为d（0≤d≤2）。当您将杯子倾斜到最大角度以致没有倒出内部液体时，液体表面的面积是多少？
思路：
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-10
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '
' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '
' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
double d;
const double pi = acos(-1);
int sgn(double x)
{
if (fabs(x) < eps) {
return 0;
} else if (x > 0) {
return 1;
} else {
return -1;
}
}
double get_V(double mid)
{
double a0 = acos(1 - 2 * tan(mid));
double res = (sin(a0) - a0 * cos(a0)) - 1.0 / 3.0 * pow(sin(a0), 3);
res /= tan(mid);
return res;
}
int main()
{
int t;
t = readint();
while (t--) {
cin >> d;
if (sgn(d) == 0) {
printf("0.00000
");
} else if (sgn(d - 1) >= 0) {
double ans = pi * sqrt(d * d - d * 4 + 5);
printf("%.5f
", ans );
} else {
double l = 0;
double r = pi / 4;
repd(rp, 1, 800) {
double mid = (l + r) * 0.5;
double res = get_V(mid);
if (sgn(res - d * pi) > 0) {
r = mid;
} else {
l = mid;
}
}
double theta = l;
double a = acos(1 - 2 * tan(theta));
double S = a - sin(a) * cos(a);
double ans = S / sin(theta);
printf("%.5f
", ans );
}
}
return 0;
}