問(wèn)題 https://www.codechef.com/problems/MATPH 的鏈接 所以，我在這個(gè)問(wèn)題上停留了幾個(gè)小時(shí)，我不知道我錯(cuò)在哪里。我使用Eratosthenes的Sieve來(lái)查找素?cái)?shù)，并將所有素?cái)?shù)保存在哈希映射中。在線法官在測(cè)試用例上給了我錯(cuò)誤的答案。        static void dri(int n) {            long large=0;int r=0,x,count=0,p,count1=0;            x=(int)Math.sqrt(n);            //To understand why I calculated x let's take an example            //let n=530 sqrt(530) is 23 so for all the numbers greater than 23 when             //we square them they will come out to be greater than n             //so now I just have to check the numbers till x because numbers             //greater than x will defiantly fail.I think you get             //what I'm trying to explain            while(r<x) {                r = map.get(++count); // Prime numbers will be fetched from map and stored in r                int exp = (int) (Math.log(n) / Math.log(r));                //To explain this line let n=64 and r=3.Now, exp will be equal to 3                //This result implies that  for r=3 the 3^exp is the //maximum(less than n) value  which I can calculate by having a prime in a power                if (exp != 1) {   //This is just to resolve an error dont mind this line                    if (map.containsValue(exp) == false) {                    //This line implies that when exp is not prime                      //So as I need prime number  next lines of code will calculate the nearest prime to exp                        count1 = exp;                        while (!map.containsValue(--count1)) ;                          exp = count1;                    }                    int temp = (int) Math.pow(r, exp);                    if (large < temp)                        large = temp;                }            }            System.out.println(large);        }