How can I print the prime numbers ending with 1 in a given range and the nearest one after b?

• Sample input :
  10
  100

• Sample output :
  11,31,41,61,71,101

• from the above code I can get the sample output value upto the value 71,how can I get nearest prime number ending with 1 after b.

Here is the code I tried:

import java.util.*;
public class Program{
public static void main(String[] args){
    Scanner in=new Scanner(System.in);
    int a=in.nextInt();
    int b=in.nextInt();
    int i,j,count;
    for(i=a;i<=b;i++)
    {
        for(j=2;j<=b;j++)
        {
            if(i%j==0)
            break;
        }
        if(j==i && j%10==1) 
        {
            System.out.println(i);
        }
    }
} 

Here's how you can do it, run the first loop indefinitely and break it when i is greater than b and it's a prime with 1 at the end.

This is the implementation just modifying your code a bit-

import java.util.*;
class Program{
    public static void main(String[] args){
      Scanner in=new Scanner(System.in);
        int a=in.nextInt();
        int b=in.nextInt();
        boolean flag = false;
        int i,j,count;
        for(i=a;;i++) {	
        	if(i>b) flag = true;
            for(j=2;j<=i;j++) {
                if(i%j==0)
                break;
            }
            if(j==i && j%10==1) {
            	System.out.println(i);
        		if(flag) break;		
            }
        }
    }
} 

Add condition in both the for loops like I<= b*2;

After that before printing the series add one more if block or add the condition in if(j==1 && j%10==1)
Condition-
You should have to check what is value of i and whether it's greater than b or not. If it's greater than b then it should be end with 1 and nearer value otherwise no need to print and break it as other values you don't need at all.

Try this with a few improvements.

• first check to see if divisible by 2
• then, since 2 is taken care of, divide only by odd numbers<br>
  starting with 3, not exceeding the sqrt of the number under test.
• then just continue until the search parameter are met.

		Scanner in = new Scanner(System.in);
		int a = in.nextInt();
		int b = in.nextInt();
		outer:
        
		for (int i = a; i <= Integer.MAX_VALUE; i++) {
		  // first check division by 2 so you can increment by 2 later
	      if (i % 2 == 0) {
				continue;
			}
            // increment from 3 to square root of i by 2's
			for (int j = 3; j <= Math.sqrt(i); j+=2) {
				if (i % j == 0) {
                    // not prime so continue with outer loop
					continue outer;
				}
			}
			if (i % 10 == 1) {
				System.out.print(i + " ");
                // keep searching until i > b.
                if (i > b) {
				   break;
                }
			}
		}

for input of 10 100 Prints

 11 31 41 61 71 101 

</details>



# 答案4
**得分**: 0

你无需将一个数字除以它前面的所有数字来判断它是否为质数。只需检查它的平方根以下的数字即可。请参阅https://en.wikipedia.org/wiki/Primality_test。操作如下：

使用 `for` 循环
===

```java
import java.util.Scanner;

public class Program {

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        System.out.print("输入两个以空格分隔的整数：");
        int a = in.nextInt();
        int b = in.nextInt();
        int i, j, sqrt;

        // 注意，在声明 'for' 循环时有三个部分：
        // for (初始化; 条件; 改变)，其中没有哪个部分是必需的。在下面给出的循环语法中没有放置终止条件。终止条件已在打印质数之后设置。
        for (i = a;; i++) {
            sqrt = (int) Math.sqrt(i);
            for (j = 2; j <= sqrt; j++) {
                if (i % j == 0) {
                    break;
                }
            }
            // 如果 j 的循环没有中断，该数字是质数。注意 1 不是质数。此外，数字的最后一位应为 1。
            if (j > sqrt && Math.abs(i) != 1 && i % 10 == 1) {
                System.out.print(i + " "); // 打印质数
                if (i >= b) { // 终止条件
                    break;
                }
            }
        }
    }
}

或者，使用 while 循环

import java.util.Scanner;

public class Program {

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        System.out.print("输入两个以空格分隔的整数：");
        int a = in.nextInt();
        int b = in.nextInt();
        int i = a, j, sqrt;

        while (true) {
            sqrt = (int) Math.sqrt(i);
            for (j = 2; j <= sqrt; j++) {
                if (i % j == 0) {
                    break;
                }
            }
            // 如果 j 的循环没有中断，该数字是质数。注意 1 不是质数。此外，数字的最后一位应为 1。
            if (j > sqrt && Math.abs(i) != 1 && i % 10 == 1) {
                System.out.print(i + " "); // 打印质数
                if (i >= b) { // 终止条件
                    break;
                }
            }
            i++;
        }
    }
}

一个示例运行：

输入两个以空格分隔的整数：10 100
11 31 41 61 71 101

另一个示例运行：

输入两个以空格分隔的整数：10 200
11 31 41 61 71 101 131 151 181 191 211

import java.util.Scanner;

public class Program {

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        System.out.print("Enter two integers separated by space: ");
        int a = in.nextInt();
        int b = in.nextInt();
        int i, j, sqrt;

        // Note that there are three sections in the declaration of a 'for' loop:
        // for(initialization;condition;change) where none of the sections is
        // mandatory. There is no condition put in the loop syntax given below. The
        // condition for termination has been put after printing the prime number.
        for (i = a;; i++) {
            sqrt = (int) Math.sqrt(i);
            for (j = 2; j <= sqrt; j++) {
                if (i % j == 0) {
                    break;
                }
            }
            // If the loop with j completed without a break, the number is prime. Note that
            // 1 is not a prime number.Also, the last digit of the number should be 1.
            if (j > sqrt && Math.abs(i) != 1 && i % 10 == 1) {
                System.out.print(i + " "); // Print the prime
                if (i >= b) {// Condition for termination
                    break;
                }
            }
        }
    }
}

import java.util.Scanner;

public class Program {

	public static void main(String[] args) {

		Scanner in = new Scanner(System.in);
		System.out.print("Enter two integers separated by space: ");
		int a = in.nextInt();
		int b = in.nextInt();
		int i = a, j, sqrt;

		while (true) {
			sqrt = (int) Math.sqrt(i);
			for (j = 2; j <= sqrt; j++) {
				if (i % j == 0) {
					break;
				}
			}
			// If the loop with j completed without a break, the number is prime. Note that
			// 1 is not a prime number.Also, the last digit of the number should be 1.
			if (j > sqrt && Math.abs(i) != 1 && i % 10 == 1) {
				System.out.print(i + " "); // Print the prime
				if (i >= b) {// Condition for termination
					break;
				}
			}
			i++;
		}
	}
}

Enter two integers separated by space: 10 100
11 31 41 61 71 101

Enter two integers separated by space: 10 200
11 31 41 61 71 101 131 151 181 191 211

答案5

得分: 0

我假设您需要满足以下条件的质数：

• 在一个范围内（从较小到较大）
• 以1结尾
• 最接近查询值

如果是这样的话，最好的策略是在两个方向（低于和高于）进行外向搜索，并在首次找到以1结尾的质数时停止。

public static int nearest1Prime(final int lower, final int upper, final int val)
{
    if(val < lower || val > upper) return 0;
    
    int before, after;
    if((val % 10) == 1) 
    {
        if(isPrime(val))
            return val;
        before = val - 10;
        after = val + 10;
    }
    else
    {
        int base = 10 * (val / 10);
        if(val == base)
        {
            after = base+1;
            before = after-10;
        }
        else
        {
            before = base+1;
            after = before+10;
        }
    }

    int prime = 0;
    
    while(prime == 0 && (before >= lower || after <= upper))
    {
        if(before >= lower && isPrime(before)) 
            prime = before;

        if(after <= upper && isPrime(after) && (prime == 0 || (after-val) < (val-before)))
            prime = after;	
		
        before -= 10;
        after -= 10;
    }
    
    return prime;
}

public static boolean isPrime(int v)
{
    for(int i=(int)Math.sqrt(v); i>1; i--)
    {
        if((v % i) == 0) return false;
    }
    return true;
}

测试：

int lower = 10;
int upper = 100;

for(int i=lower; i<=upper; i++)
{
    int prime = nearest1Prime(lower, upper, i);
    System.out.println("Nearest Prime: " + i + " : " + prime);
}

输出：

10 : 11
11 : 11
12 : 11
13 : 11
14 : 11
15 : 11
16 : 11
17 : 11
18 : 11
19 : 11
20 : 11
21 : 11
22 : 31
23 : 31
24 : 31
25 : 31
26 : 31
27 : 31
28 : 31
29 : 31
30 : 31
31 : 31
32 : 31
33 : 31
34 : 31
35 : 31
36 : 31
37 : 41
38 : 41
39 : 41
40 : 41
41 : 41
42 : 41
43 : 41
44 : 41
45 : 41
46 : 41
47 : 41
48 : 41
49 : 41
50 : 41
51 : 41
52 : 61
53 : 61
54 : 61
55 : 61
56 : 61
57 : 61
58 : 61
59 : 61
60 : 61
61 : 61
62 : 61
63 : 61
64 : 61
65 : 61
66 : 61
67 : 71
68 : 71
69 : 71
70 : 71
71 : 71
72 : 71
73 : 71
74 : 71
75 : 71
76 : 71
77 : 71
78 : 71
79 : 71
80 : 71
81 : 71
82 : 71
83 : 71
84 : 71
85 : 71
86 : 71
87 : 71
88 : 71
89 : 71
90 : 71
91 : 71
92 : 71
93 : 71
94 : 71
95 : 71
96 : 71
97 : 71
98 : 71
99 : 71
100 : 71

public static int nearest1Prime(final int lower, final int upper, final int val)
{
if(val < lower || val > upper) return 0;
int before, after;
if((val % 10) == 1) 
{
if(isPrime(val))
return val;
before = val - 10;
after = val + 10;
}
else
{
int base = 10 * (val / 10);
if(val == base)
{
after = base+1;
before = after-10;
}
else
{
before = base+1;
after = before+10;
}
}
int prime = 0;
while(prime == 0 && (before >= lower || after <= upper))
{
if(before >= lower && isPrime(before)) 
prime = before;
if(after <= upper && isPrime(after) && (prime == 0 || (after-val) < (val-before)))
prime = after;	
before -= 10;
after -= 10;
}
return prime;
}
public static boolean isPrime(int v)
{
for(int i=(int)Math.sqrt(v); i>1; i--)
{
if((v % i) == 0) return false;
}
return true;
}

int lower = 10;
int upper = 100;
for(int i=lower; i<=upper; i++)
{
int prime = nearest1Prime(lower, upper, i);
System.out.println("Nearest Prime: " + i + " : " + prime);
}

10 : 11
11 : 11
12 : 11
13 : 11
14 : 11
15 : 11
16 : 11
17 : 11
18 : 11
19 : 11
20 : 11
21 : 11
22 : 31
23 : 31
24 : 31
25 : 31
26 : 31
27 : 31
28 : 31
29 : 31
30 : 31
31 : 31
32 : 31
33 : 31
34 : 31
35 : 31
36 : 31
37 : 41
38 : 41
39 : 41
40 : 41
41 : 41
42 : 41
43 : 41
44 : 41
45 : 41
46 : 41
47 : 41
48 : 41
49 : 41
50 : 41
51 : 41
52 : 61
53 : 61
54 : 61
55 : 61
56 : 61
57 : 61
58 : 61
59 : 61
60 : 61
61 : 61
62 : 61
63 : 61
64 : 61
65 : 61
66 : 61
67 : 71
68 : 71
69 : 71
70 : 71
71 : 71
72 : 71
73 : 71
74 : 71
75 : 71
76 : 71
77 : 71
78 : 71
79 : 71
80 : 71
81 : 71
82 : 71
83 : 71
84 : 71
85 : 71
86 : 71
87 : 71
88 : 71
89 : 71
90 : 71
91 : 71
92 : 71
93 : 71
94 : 71
95 : 71
96 : 71
97 : 71
98 : 71
99 : 71
100 : 71