p144.ans1
======================
0.5555556

=================
p144.cpp
======================
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream &out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream &in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream &in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream &out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

double X, Y, Z;

void read() { cin >> X >> Y >> Z; }

void solve() {
    // Both arrival times are uniform over a window of delta = (Y - X) * 60
    // minutes, so the sample space is a delta x delta square. They meet when
    // their arrival times differ by at most Z, the diagonal band
    // |t1 - t2| <= Z. Its area is delta^2 minus the two corner triangles of
    // total area (delta - Z)^2, and we report that area divided by delta^2,
    // written here as Z^2/delta^2 + 2*(delta - Z)*Z/delta^2.

    double delta = (Y - X) * 60;
    double ans = (Z * Z) / (delta * delta);
    ans += 2 * (delta - Z) * Z / (delta * delta);
    cout << fixed << setprecision(7) << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}

=================
p144.in1
======================
11 12 20.0

=================
statement.txt
======================
144. Meeting

time limit per test: 0.25 sec.
memory limit per test: 4096 KB


Two of the three members of the winning team of one of the ACM regional contests are going to meet in order to train for the upcoming World Finals. They decided that they will meet sometime between X o'clock and Y o'clock. Because they never get anywhere on time (they were late even on the day of the regional contest), they did not set an exact time when they will meet. However, they decided that the one who gets first at the meeting point will not wait more than Z minutes for the other one (they calculated that, if the other one will not come within Z minutes from the arrival of the first of them, then it is very probable that he will not show up at all).
Knowing that, in the end, both of them will show up at some time between X o'clock and Y o'clock (not necessarily after an integer number of minutes), compute which is the probability that they will actually meet.


Input

The input will contain 2 integer numbers X and Y (0<=X<Y<=24) and one real number Z ( 0 < Z <= 60*(Y-X) ).


Output

You should output the required probability with 7 decimal digits (rounded according to the 8th decimal digit).


Sample Input

11 12 20.0
Sample Output

0.5555556
Author	: Mugurel Ionut Andreica
Resource	: SSU::Online Contester Fall Contest #2
Date	: Fall 2002

=================
