[英]Given a number n, return true is all the factors of n are prime numbers. Note that 1 and the number itself are not considered as factors
public class AllFactorsArePrime {
public static void main(String[] args) {
AllFactorsArePrime obj = new AllFactorsArePrime();
boolean result = obj.areAllFactorsPrime(8);
System.out.println(result);
}
public boolean areAllFactorsPrime(int n) {
int j=0;
double k=n;
while(n%2==0){
n=n/2;
j=2;
}
for(int i=3; i<=n;i=i+2){
while(n%i==0){
n=n/i;
j=i;
}
}
if(j==0 ){
return 1;
}
return j;
}
上面的代碼返回了主要因素,但是返回應該是對還是錯。有什么建議嗎? 樣本輸入#1
areAllFactorsPrime(22)
樣本輸出#1
true
樣本輸入2
areAllFactorsPrime(25)
樣本輸出2
true
樣本輸入#3
areAllFactorsPrime(32)
樣本輸出3
false
我認為這就是您要實現的目標!
import java.util.*;
import java.lang.*;
import java.io.*;
public class AllFactorsArePrime {
public static void main(String[] args) {
AllFactorsArePrime obj = new AllFactorsArePrime();
boolean result = obj.areAllFactorsPrime(32);
System.out.println(result);
}
public boolean areAllFactorsPrime(int n) {
int count=0;
int j=0;
double k=n;
while(n%2==0){
n=n/2;
j=2;
count++;
}
for(int i=3; i<=n;i=i+2){
while(n%i==0){
n=n/i;
j=i;
count++;
}
}
if(count>=3)
{
return false;
}
else
return true;
}
}
如果除1之外有2個以上素數,則邏輯很簡單,這意味着您至少有1個復合素,即prime1 * prime2。
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