問題 https://www.codechef.com/problems/MATPH 的鏈接 所以,我在這個問題上停留了幾個小時,我不知道我錯在哪里。我使用Eratosthenes的Sieve來查找素數,并將所有素數保存在哈希映射中。在線法官在測試用例上給了我錯誤的答案。 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); }
1 回答

慕慕森
TA貢獻1856條經驗 獲得超17個贊
對于每個測試用例,在包含最大漂亮數字的一行中輸出≤ N。如果不存在這樣的數字,則打印 ?1。
我相信4是最小的美麗數,因為2是最小的素數,2^ 2等于4。N 只需要≥ 0。所以, ,并且應該全部打印。我試過了。他們沒有。我相信這就是你在CodeChef上失敗的原因。dri(0)
dri(1)
dri(2)
dri(3)
?1
我把它留給你自己來了解提到的對你的方法的調用是如何表現的,以及如何處理它。
順便說一句,將素數保存在地圖中有什么意義?列表或排序集不是更合適嗎?
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