Problem-solving ideas : After reading the real problem analysis, I think the method he gave is more cumbersome, that is, after each deletion, the elements of the back array are moved forward, and the upper and lower limits of the sentinel and pointer must be set accurately. In the question, The method used is to directly assign a value of 0 to the element to be deleted, which is equivalent to a placeholder. The subsequent selection only needs to be conditional, and these deleted values can be ignored.
Difficulty in solving the problem : This problem is still a pitfall. The example given in the problem allows me to shift, and the lucky number will be cut off when the lucky number reaches 7, and then get the same sequence of numbers as his, and then filter and compare it. I didn’t expect people to want it. The lucky number filtering is carried out until the filtering is meaningless (that is, the filtering interval exceeds the length of the array), so the code is changed again!
package 蓝桥杯;
import java.util.Scanner;
public class 省_幸运数 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
//定义范围上下限
int n = in.nextInt();
int m = in.nextInt();
int[] arr = new int[m/2];
//先做一轮筛选，将数组都填上奇数
for (int i = 0; i < m/2; i++)
arr[i] = i*2+1;
//开始依次进行幸运数筛选
int k = 1;
while (true){
//arr[K]中存放的即为幸运数，当他的大小超过数组长度，就可以结束循环
if(arr[k] > m/2)
break;
//设置计数器
int add = 0;
//遍历依次筛选幸运数
for (int i = 0; i < m/2; i++) {
if(arr[i] == 0)
continue;
add++;
if(add % arr[k] == 0){
add = 0;
arr[i] = 0;
}
}
k++;
while (arr[k] == 0) k++;
}
//找到上限指针
int flag = 0;
for (int i = 0; i < m/2; i++) {
if(arr[i] >= n){
flag = i;
break;
}
}
//从上限开始依次遍历累加，计算幸运数个数
int add = 0;
for (int i = flag; i < m/2; i++) {
if(arr[i] == 0 || arr[i] == n)
continue;
else{
if(arr[i] >= m)
break;
add++;
}
}
System.out.print(add);
}
}