我尝试编写BiggerThanPrime程序,该程序允许用户给出输入p,程序可以找到下一个最接近的素数(n),使得(n-p)的值最小。
这是我的代码:
static boolean Prime (int n){
boolean NotPrime = true;
boolean result = true;
for (int i=2; i< n; i++){
NotPrime = (n % i != 0);
result = (result && NotPrime);
}
return result;
}
//Using Betrand's Postulate which states that there always exists at least one prime p s.t.a< p <2a
public static void main(String[] args) {
int p = Integer.parseInt(args[0]);
int k = p+1;
while( k > p && k< 2*p){
if( Prime(k) == true){
System.out.println("the next bigger prime than "+ p + " is "+ k);
} else{
k++;
}
}
}
但是while循环进入无限循环。
结果是:
the next bigger prime than 20 is 23
the next bigger prime than 20 is 23
.
.
.
(infinitely goes on) :(
我做错了什么?
答案 0 :(得分:2)
一旦获得更大的素数,你需要打破循环,并且你需要在每次迭代中增加k,如下所示:
while( k > p && k< 2*p){
if(Prime(k)){
System.out.println("the next bigger prime than "+ p + " is "+ k);
break;
}
k++;
}
还请注意,当Prime(k)返回一个布尔值时,不必再将它与true进行比较。
答案 1 :(得分:1)
一旦找到下一个素数,就应该从循环中断开:
while( k > p && k< 2*p){
if( Prime(k) == true){
System.out.println("the next bigger prime than "+ p + " is "+ k);
break;
} else{
k++;
}
}
答案 2 :(得分:1)
因为您没有在每次迭代时递增k
试试这个:
while( k > p && k< 2*p){
if( Prime(k) == true){
System.out.println("the next bigger prime than "+ p + " is "+ k);
}
k++;
}
答案 3 :(得分:1)
while( k > p && k< 2*p){
if( Prime(k) == true){
System.out.println("the next bigger prime than "+ p + " is "+ k);
} else{
k++;
}
}
一旦Prime(k)返回true,你就不会改变k的值,因此循环继续。
答案 4 :(得分:1)
当
( Prime(k) == true)
时,它永远不会增加k的值。所以 它将进入无限循环。
你可以用这两种方式改变lop
方式1:
while( k > p && k< 2*p){
if( Prime(k) == true){
System.out.println("the next bigger prime than "+ p + " is "+ k);
}
k++;
}
方式2:
while( k > p && k< 2*p){
if( Prime(k) == true){
System.out.println("the next bigger prime than "+ p + " is "+ k);
Break;
} else{
k++;
}
}
答案 5 :(得分:0)
import java.util.*;
public class index_page
{
public static void main(String[] args) {
Scanner keyB = new Scanner(System.in);
String code = keyB.next();
while (!code.equals("01") || code.equals("02") || code.equals("00") || code.equals("03") || code.equals("04") || code.equals("05")){//Checking for erros in code from user
System.out.println("Invalid code!! Try again");
code = keyB.next();
}
}
}