UVa/10573 - Geometry paradox/10573.cpp

   1 /*
   2   Problem: 10573 - Geometry Paradox (UVa)
   3  */
   4 using namespace std;
   5 #include <algorithm>
   6 #include <iostream>
   7 #include <iterator>
   8 #include <sstream>
   9 #include <fstream>
  10 #include <cassert>
  11 #include <climits>
  12 #include <cstdlib>
  13 #include <cstring>
  14 #include <string>
  15 #include <cstdio>
  16 #include <vector>
  17 #include <cmath>
  18 #include <queue>
  19 #include <deque>
  20 #include <stack>
  21 #include <list>
  22 #include <map>
  23 #include <set>
  24
  25 #define foreach(x, v) for (typeof (v).begin() x = (v).begin(); x != (v).end(); ++x)
  26 #define D(x) cout << #x " is " << x << endl
  27 #define For(i, a, b) for (int i=(a); i<(b); ++i)
  28
  29 int main(){
  30   int n;
  31   cin >> n;
  32   string s;
  33   getline(cin, s);
  34   while (n-- && getline(cin, s)){
  35     int r1, r2;
  36     if (sscanf(s.c_str(), "%d %d", &r1, &r2) == 2){
  37       printf("%.4lf\n", 4*acos(0)*r1*r2);
  38     }else{
  39       int t;
  40       sscanf(s.c_str(), "%d", &t);
  41       printf("%.4lf\n", acos(0)*t*t/4);
  42     }
  43   }
  44   return 0;
  45 }
  46
  47 /*
  48   If you are given r1 and r2, then the gray area is
  49   the area of the big circle minus the area of the small circles:
  50
  51   pi * ((r1 + r1)² - r1² - r2²)) = 2 * pi * r1 * r2
  52
  53   If you are given t, you can form a rectangular triangle where the cathetuses
  54   are (t/2) and k, and the hypothenuse is (r1 + r2). Then, k = (r1 - r2).
  55   Solve the pythagorean theorem for this triangle and you should get
  56
  57   r1 * r2 = t² / 16
  58
  59   Replace r1 * r2 in the formula above and you will get
  60   2 * pi * r1 * r2 = 2 * pi * (t² / 16) = pi * t² / 8
  61  */