Problem description
Enter a positive integer n, and determine how many times the number 1 should appear from 1 to n. For example, for the number 1123, 1 appears twice. For example, 15, from 1 to 15, there are a total of 8 1s.
Input format
a positive integer n
output format
an integer, representing the data that 1 appears.
Sample input
15
Sample output
8
Data size and agreement
n does not exceed 30000
Solution:
The first reaction to this question is to first convert the number into a string, and judge whether each number in the string is the string "1". If it is, add one to temp. code show as below:
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner scanner=new Scanner(System.in);
int n=scanner.nextInt();
scanner.close();
int temp=0;
for(int i=1;i<=n;i++) {
//从1循环到n
String string=i+"";
for(int j=0;j<string.length();j++) {
//当是两位数及以上时，就要对每位上的数进行判断
char a=string.charAt(j);
boolean status=(a+"").contains("1");
if(status) {
temp++;
}
}
}
System.out.println(temp);
}
}
Evaluation result:
Although the result passed the evaluation, but I always felt that my method was not very good, so I went to find the way of the big guy. Sure enough, the boss's method is concise and clear.
Solution:
The following code is mainly to judge whether the number 1 is included by performing the remainder and rounding operations of 10 for each number.
The code is as follows (Address: https://blog.csdn.net/a1439775520/article/details/104214711):
import java.util.Scanner;
public class 一的个数 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
//使用完输入流要及时得关闭，防止占内存
sc.close();
//先把所有得变量都声明了
int count = 0,a,b;
for (int i = 1; i <= n; i++) {
//用一个变量代替，如果更改i就会更改循环
a=i;
//和水仙花数一样得方法，取最后一位，/10删除最后一位
while (a!=0){
b=a%10;
if (b==1) count++;
a/=10;
}
}
System.out.println(count);
}
}
Evaluation results:
