问题链接-https://www.codechef.com/problems/MATPH 因此,我在这个问题上停留了几个小时,而且我不知道自己错在哪里。 我使用Eratosthenes的Sieve来查找素数,并将所有素数保存在哈希图中。在线法官给我关于测试用例的错误答案。
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);
}
我
答案 0 :(得分:0)
对于每个测试用例,在一行中包含最大的输出 美丽数字≤ N 。如果没有这样的数字,则打印-1。
我相信4是最小的美丽数字,因为2是最小的质数,并且2 ^ 2等于4。 N 仅需≥0。所以dri(0)
,{{ 1}},dri(1)
和dri(2)
应该全部打印dri(3)
。我试过了。他们没有。我相信这就是您在CodeChef上失败的原因。
我将自己留给自己,以了解所提及的对您的方法的调用的行为以及处理方法。
顺便说一句,将质数保存在地图中有什么意义?列表或排序集会更合适吗?