There is multiple test cases , in each test case there is only one line contains a number N(2<=N<=100000). Process to the end of file. Output For each test case , output one line contains one number , indicating different prime factor in the number N. Sample Input 216124123512346 Samp...
Now, your task is telling me what position of the largest prime factor. The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc. Specially, LPF(1) = 0. Input Each line will contain one integer n(0 < n < 1000000). Output Output the LPF(n). Sample Input 1 2 3...
【摘要】 Problem Description Everybody knows any number can be combined by the prime number. Now, your task is telling me what position of the largest prime factor. The position of prime 2 is ... Problem Description Everybody knows any number can be combined by the prime number. Now, you...
util.HashSet; import java.util.Scanner; import java.util.Set; /** * Program to find out prime factors of given number * @author includehelp */ public class PrimeFactors { /** * method to find prime factor for supplied number * @param number * @return */ static String getPrimeFactors...
简介:HDOJ(HDU) 2136 Largest prime factor(素数筛选) Problem Description Everybody knows any number can be combined by the prime number. Now, your task is telling me what position of the largest prime factor. The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc. ...
The largest prime factor(最大质因数) 1. 问题: The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? 2. 解法(by java in Eclipse) way1 way2 注:利用分解质因数的方法,从小向大用质数整除(如果此质数恰好是n的因数的话)n,即不断的...
for(intx =2; x < (int) sqrtf((float) num) +1; x++)/* go through numbers up to the number's square root looking for a factor */ { if(num % x ==0) { returnfalse;/* has a factor, so not a prime */ } } returntrue;/* if we've made it this far it's a prime */...
Factor N in the format N = p1 ^k1 *p2 ^k2 *…*pm ^km , where pi 's are prime factors of N in increasing order, and the exponent ki is the number of pi – hence when there is only...
= num ); if( allPrime ) System.out.println("It is Circular Prime number." ); else System.out.println("It is not a Circular Prime number." ); } public static boolean isPrime( int n ){ int factorCount = 0; if( n < 2 ) return false; else if( n == 2 ) return true; else...
Now, your task is telling me what position of the largest prime factor. The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc. Specially, LPF(1) = 0. Input Each line will contain one integer n(0 < n < 1000000). ...