## Algorithm

Problem Statement for BirthdayOdds

### 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) (be sure your method is public)

### 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`

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>

using namespace std;

class BirthdayOdds {
public:
int minPeople(int minOdds, int DaysInYear) {
double comp = 100 - minOdds;
double all = 100;
for(int i = DaysInYear - 1; i >= 0; --i) {
if(all <= comp)
return DaysInYear - i;
all *= (1.0 * i / DaysInYear);
}
return DaysInYear + 1;
}
};``````
Copy The Code &

Input

cmd
50
365

Output

cmd
23