Hi I think that we should help each other. If someone resolves a problem he/she must help the others with materials or links where the algorithm/formula is explained. This is the reason we are users of such sites - because we want to improve our programming skills.
So I decided to help all of you who find this problem difficult. Here you can find help:

http://en.wikipedia.org/wiki/Circular_segment

http://www.mathopenref.com/segmentarea.html

The rest is up to you!
HTH

Why I have WA?

#include <iostream>
#include <cmath>

using namespace std;

int main() {
    double const pi = 3.14159265358979323846;
    double a;
    double r;
    cin >> a >> r;
    long double s = pi * r * r;
    if (r <= a / 2) {
        cout << roundl(s * 1000) / 1000;
    } else if (r >= ((a / 2) * sqrt(2))) {
        cout << (int) (a * a) << endl;
    } else {
        double cosx;
        cosx = (a / 2) / r;
        double rad;
        rad = acos(cosx) * 2;
        double segm = (r * r / 2) * (rad - sin(rad));
        s = s - 4 * segm;
        cout << round(s * 1000) / 1000 << endl;
    }
    return 0;
}

#include <bits/stdc++.h>

#define gif(a,b) ((a)/(b)+((a)%(b)?1:0))
#define float long double
#define int long long
//#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define w(x) cout<<(#x)<<" is "<<(x)<<"\n";
using namespace std;

int32_t main()
{
    //IOS
    float a, r;
    cin >> a >> r;
    float theta = acos(a/(2*r));


    float s = sin(theta);

    float pi = acos(-1);
    if(a/2 > r)
        printf("%0.3Lf",pi*r*r);
    else if(r > a/pow(2, 0.5))
        cout << a*a;
    else
    cout << (pi*r*r) - 4*(r*r*theta - a*r*s/2) ;
    return 0;
}

#include<iostream>
#include<cmath>
#include<iomanip>
#define pi 3.1415926
using namespace std;

int main(){
    float side,len;
    cin>>side>>len;
    if(len>(side/2)*sqrt(2)){
        cout<<fixed<<setprecision(3)<<side*side;
        return 0;
    }
    if((side/2)>=len){
        cout<<fixed<<setprecision(3)<<len*len*pi;
        return 0;
    }
    float cosx=(side/2)/len;
    float sinx=sqrt(1-cosx*cosx);
    float cosA=2*sinx*cosx;
    float theta=acos(cosA);
    cout<<fixed<<setprecision(3)<<len*len*0.5*theta*4+sqrt(len*len-(side/2)*(side/2))*(side/2)*4;

return 0;
}

//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("avx")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

#define re return
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define sqrt(x) sqrt(abs(x))
#define mp make_pair
#define pi (3.14159265358979323846264338327950288419716939937510)
#define fo(i, n) for(int i = 0; i < n; ++i)
#define ro(i, n) for(int i = n - 1; i >= 0; --i)
#define unique(v) v.resize(unique(all(v)) - v.begin())

template <class T> T abs (T x) { re x > 0 ? x : -x; }
template <class T> T sqr (T x) { re x * x; }
template <class T> T gcd (T a, T b) { re a ? gcd (b % a, a) : b; }
template <class T> int sgn (T x) { re x > 0 ? 1 : (x < 0 ? -1 : 0); }

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef double D;
typedef long double ld;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef unsigned long long ull;
typedef tree <pair<int, char>, null_type, less<pair<int, char>>, rb_tree_tag, tree_order_statistics_node_update> _tree;

int main() {
    cout.precision(10);
    double r, a;
    cin >> a >> r;
    double h = 2 * sqrt(r * r - a * a / 4);
    if (r <= a / 2) cout << pi * r * r / 2 << endl;
    else if (r >= a * sqrt(2) / 2) cout << a * a << endl;
    else cout << pi * r * r + a * h - 2 * r * r * (double)asin(a * h / (2 * r * r)) << endl;
    re 0;
}

I have WA, when I defined pi with 5 numbers after point.
and when I have tried pi = 3.1415926535, I got AC.

It's strange a little)

if(a/2. >= r)
{
    printf("%.3lf", pi*r*r);
    return 0;
}
This is 4th test, use only "%.3lf" !

I'm getting wrong ans on test case 3. Can someone please give a hint?

Here's my solution in Java:

FastReader sc = new FastReader(System.in);
        double a = sc.nextDouble();
        double r = sc.nextDouble();
        double ans;

        if (r <= a / 2) {
            ans = Math.PI * r * r;
        } else {
            double angle = (2 * Math.PI) - (8 * Math.acos(a / (2 * r)));
            ans = (r * r * angle / 2) + (2 * a * Math.sqrt(r * r - a * a / 4));
        }

        System.out.println(String.format("%.3f", ans));

Maybe you can try to declare l,r as double not int  :)

Simple but interesting problem. There are two ways to find angle: geometrically (i made it this way) or computationally :).

Может я чего-то не догоняю, но у меня всё работает на компе и высчитывает вроде верно, а при заливе кода на сайт пишет ошибку компиляции. Прошу помощи более опытных

#include <iostream>
#include <math.h>
using namespace std;

int main()
{
    const double pi = 3.141592653589793238462643383279;
    int n = 0, r = 0;
    double k = 0, s = 0, x = 0, ot = 0, d = 0, a = 0,acos = 0;
    cin >> n >> r;
    a = double (sqrt(2 * n*n)) / 2; /*Нахождение диагонали квадрата, для сравнения с R*/
    if (r <= n/2) ot = pi * r*r;
    if (r >= a) ot = n * n;
    else
    {
        d = n / 2;
        acos = (pi / 2) - atan((d / r) / sqrt(1 - (d / r)*(d / r)));
        x = 2 * acos;
        s = 4 * (0.5*r*r*(x - sin(x)));
        ot = pi*r*r - s;
    }
    ot = round(ot * 1000) / 1000;
    cout << ot;
    system("pause");
    return 0;
}

не фига ответ не такой 95.0.... я посчитал другой пусть нормально пишут

thanks...

#include <stdio.h>
#include <math.h>

#define PI 3.1416

int main()
{
        double a, r, area;

        scanf("%lf%lf", &a, &r);
        a /= 2;

        if (a >= r) {
                printf("%.3f\n", PI * r * r);
                return 0;
        }
        if (r >= sqrt(2) * a) {
                printf("%.3f\n", 4 * a * a);
                return 0;
        }
        printf("%.3f\n", PI * r * r - 4 * (acos(a / r) * r * r - a *
sqrt(r * r - a * a)));
        return 0;
}

var a,r:integer;  az,s,stru,b:real;
function ArcCos ( X : Real ): Real;
var
tmpArcCos : Real;
begin
if X = 0.0 then
tmpArcCos := Pi / 2.0
else
tmpArcCos := ArcTan ( Sqrt ( 1 - X * X ) / X );
if X < 0.0 then
tmpArcCos := Pi - tmpArcCos;
ArcCos := 180*tmpArcCos/pi;
end;


begin
readln(a,r);
if a/2>=r then writeln(pi*r*r:1:3) else
if sqrt(2)/2*a<=r then writeln(a*a) else
 begin
 az:=2*arccos((a/2)/r);
   s:=pi*r*r*az/360;
   b:=2*sqrt((r-a/2)*(r+a/2));
   stru:=a/4*b;
   s:=s-stru;
   s:=pi*r*r-4*s;
   writeln(s:1:3);
 end;
readln;
end.

#include <iostream>
#include <cmath>
#include <stdio.h>
int main()
{
    using namespace std;
    int nRadius, nLenght;
    double dAnswer, dPi = 3.141592654;
    cin >> nLenght >> nRadius;
    if (nRadius * nRadius >= nLenght * nLenght / 2) {
        dAnswer = nLenght * nLenght;
    } else if (nRadius <= nLenght / 2) {
        dAnswer = nRadius * nRadius * dPi;

    } else {
        double dX = sqrt(nRadius * nRadius - nLenght * nLenght / 4);
        double dSg = dPi * nRadius * nRadius / 360 * (360 - 8 * atan2(dX, nLenght / 2) *  57.296);
        dAnswer = dSg + 4 * nLenght / 2 * dX;
    }
    printf("%0.3f", dAnswer);

    return 0;
}

Edited by author 31.08.2013 17:00

Edited by author 31.08.2013 17:00

#include <iostream>
#include <fstream>
#include <cmath>
#define PI 3.14159265358979323846
using namespace std;

#ifndef ONLINE_JUDGE
    #define cin fin
    ifstream fin("input");
#endif

double N, R;

int main() {
    cin >> N >> R;
    N /= 2.0;

    if (R <= N) {
        cout << PI * R * R << '\n';
        return 0;
    }
    if (R >= N * sqrt(2.0)) {
        cout << 4.0 * N * N << '\n';
        return 0;
    }

    double A, B, C;
    A = N;
    C = R;
    B = sqrt(C * C - A * A);
    double angle = 90.0 - (acos(A / C) * 180.0 / PI) * 2.0;

    printf("%.3lf\n", 4.0 * (A * B + PI * R * R * angle / 360.0));
}

This program works fine on all the tests I saw on the forum. But I get WA #1 probably because of the wrong precision. I printed the value with 4 and 6 decimals but I get the same WA #1.
Could anyone tell me please how can I solve the "error" ? It's annoying to have a good solution and take WA.
Thanks a lot!


LATER EDIT:
It seems I succeeded with such a function:

inline void precise(double X, const int &decimals) {
    cout << (int)(X);
    cout << '.';
    X = X - (int)(X);
    for (int i = 1; i <= decimals; ++i) {
        X = X * 10.0;
        cout << (int)(X);
        X = X - (int)(X);
    }
    cout << '\n';
}

...

precise((double)(4.0 * (A * B + PI * R * R * angle / 360.0)), 3);

Have fun!

Edited by author 25.01.2014 04:19

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <cstdlib>
#include <math.h>
#define PI 3.141592653589793238462643383279502884197169399375
using namespace std;
int main(int argc, char** argv){
    double a, l, angle = 0.0;
    cin >> a >> l;
    if (l <= (a/2.0))
        printf("%0.3f", (l*l*PI));
    else if (l >= (a*sqrt(2.0)/2.0))
        printf("%0.3f", (a*a));
    else {
        while (l*cos(angle) > (a/2.0)) angle += 0.0000001;
        printf("%0.3f", 4.0*((a/2.0)*l*sin(angle)+(l*l*(PI/4.0-angle))));
    }
}

Use pi = 3.141592654

please whate is the case # 4???????????

this is my programm (WA 9):
#include <iostream>
#include <cmath>
#include <iomanip>
#define PI 3.1415926535897932
using namespace std;

int answer(int n, int k);
int main()
{
    int a,r;
    cin>>a>>r;
    double d=a/2;
    double O=2*acos(d/r);
    double s=2*r*r*(O-sin(O));
    double sk=PI*r*r;
    double sq=a*a;
    if (r<=a/2) cout<<setiosflags(ios::fixed)<<setprecision(3)<<sk<<endl;
    else if (r>=a*(sqrt(2.0)/2)) cout<<setiosflags(ios::fixed)<<setprecision(3)<<sq<<endl;
    else cout<<setiosflags(ios::fixed)<<setprecision(3)<<sk-s<<endl;
}