458. Poor Pigs

There are 1000 buckets, one and only one of them contains poison, the rest are filled with water. They all look the same. If a pig drinks that poison it will die within 15 minutes. What is the minimum amount of pigs you need to figure out which bucket contains the poison within one hour.

Answer this question, and write an algorithm for the follow-up general case.

Follow-up:

If there are n buckets and a pig drinking poison will die within m minutes, how many pigs (x) you need to figure out the “poison” bucket within p minutes? There is exact one bucket with poison.
题目解析：有1000个装了水的水桶，其中有个一有毒，一只猪喝了水以后15分钟会死亡，那么一个小时内找到下毒的水桶，最少死亡的猪，（~pig好可怜T.T~）
代码解析：如果有5桶水，那么只需要一只猪就可以找到（60min/15 = 4）也就是说可以分成（0,15,30,45,60）五个时刻，每个时刻让猪喝下一个桶里的水，就可以找到了.
如果有25桶水水桶编号成二维数组（00,01,02,03,04；10,11,12,13,14；20,21,22,23,24；30,31,32,33,34；40,41,42,43,44；）第一只猪第一时刻喝下第一位为0的水，第二只猪喝下第二位为0的水；第一只猪第二时刻喝下第一位为1的水，第二只猪同一时刻喝下第二位为1的水。以此类推可以找到有毒的水。
5^x>=1000 x=5

public class Solution {    public static void main(String[] args) {        // TODO Auto-generated method stubSystem.out.println(poorPigs(5, 15, 60));    }    public static int poorPigs(int buckets, int minutesToDie, int minutesToTest) {        if(buckets == 1){            return 0;        }        int times = minutesToTest/minutesToDie+1;        int i = 1;        int mod = 1;        while(true){            mod = mod * times;                      if(mod>= buckets){                return i;            }            i++;        }          }}