srm 174 div 1 250

Author: keshavnandan

BirthdayOdds.cpp

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+#include <iostream>
+#include <sstream>
+#include <vector>
+#include <algorithm>
+#include <set>
+#include <map>
+#include <cstring>
+#include <climits>
+using namespace std;
+typedef pair<int,int> pi;
+typedef vector<int> vi;
+typedef vector<vi> vvi;
+typedef vector<string> vs;
+typedef vector<vs> vvs;
+
+class BirthdayOdds {
+        public:
+        int minPeople(int minOdds, int daysInYear)
+        {
+            int n = daysInYear;
+            double pr = 100 - minOdds;
+            double p = 100.00;
+            int k = 1;
+            while(p > pr){
+                p *= (double)(n-k)/double(n);
+                k++;
+            }
+            return k;
+        }
+
+// BEGIN CUT HERE
+	public:
+	void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
+	private:
+	template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
+	void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
+	void test_case_0() { int Arg0 = 75; int Arg1 = 5; int Arg2 = 4; verify_case(0, Arg2, minPeople(Arg0, Arg1)); }
+	void test_case_1() { int Arg0 = 50; int Arg1 = 365; int Arg2 = 23; verify_case(1, Arg2, minPeople(Arg0, Arg1)); }
+	void test_case_2() { int Arg0 = 1; int Arg1 = 365; int Arg2 = 4; verify_case(2, Arg2, minPeople(Arg0, Arg1)); }
+	void test_case_3() { int Arg0 = 84; int Arg1 = 9227; int Arg2 = 184; verify_case(3, Arg2, minPeople(Arg0, Arg1)); }
+
+// END CUT HERE
+
+};
+
+// BEGIN CUT HERE
+     int main()
+     {
+        BirthdayOdds ___test;
+        ___test.run_test(-1);
+     }
+// END CUT HERE

BirthdayOdds.txt

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+PROBLEM STATEMENT
+Here is an interesting factoid:  "On the planet Earth, if there are at least 23 people in a room, the chance that two of them have the same birthday is greater than 50%."  You would like to come up with more factoids of this form.  Given two integers (minOdds and daysInYear), your method should return the fewest number of people (from a planet where there are daysInYear days in each year) needed such that you can be at least minOdds% sure that two of the people have the same birthday. See example 0 for further information.
+
+DEFINITION
+Class:BirthdayOdds
+Method:minPeople
+Parameters:int, int
+Returns:int
+Method signature:int minPeople(int minOdds, int daysInYear)
+
+
+NOTES
+-Two people can have the same birthday without being born in the same year.
+-You may assume that the odds of being born on a particular day are (1 / daysInYear).
+-You may assume that there are no leap years.
+
+
+CONSTRAINTS
+-minOdds will be between 1 and 99, inclusive.
+-daysInYear will be between 1 and 10000, inclusive.
+-For any number of people N, the odds that two people will have the same birthday (in a room with N people, on a planet with daysInYear days in each year) will not be within 1e-9 of minOdds. (In other words, you don't need to worry about floating-point precision for this problem.)
+
+
+EXAMPLES
+
+0)
+75
+5
+
+Returns: 4
+
+We must be 75% sure that at least two of the people in the room have the same birthday. This is equivalent to saying that the odds of everyone having different birthdays is 25% or less.
+
+If there is only one person in the room, the odds are 5/5 or 100% that nobody shares a birthday.
+If there are two people in the room, the odds are 5/5 * 4/5 = 80% that nobody shares a birthday. This is because the second person has 4 "safe" days on which his birthday could fall, out of 5 possible days in the year.
+If there are three people in the room, the odds of no overlap are 5/5 * 4/5 * 3/5 = 48%.
+If there are four people in the room, the odds are 5/5 * 4/5 * 3/5 * 2/5 = 19.2%. This means that you can be (100% - 19.2%) = 80.8% sure that two or more of them do, in fact, have the same birthday.
+
+We only need to be 75% sure of this, which was untrue for three people but true for four.  Therefore, your method should return 4.
+
+1)
+50
+365
+
+Returns: 23
+
+The factoid from the problem statement. If there are 22 people in a room, the odds of a shared birthday are roughly 47.57%. With 23 people, these odds jump to 50.73%, which is greater than or equal to the 50% needed.
+
+2)
+1
+365
+
+Returns: 4
+
+Another example from planet Earth. The odds of a repeat birthday among only four people are roughly 1.64%.
+
+3)
+84
+9227
+
+Returns: 184