输入：

每行一个正整数(<2^32).

输出：

每行一个输出,对应该数的各位数之和.

方法一：利用整数取余求解

import java.util.Scanner;
import java.lang.Math;
public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);while(sc.hasNext()) {long a = sc.nextLong();int sum = 0;//用来求和while(a!=0) {sum += a%10;a /= 10;}System.out.println(sum);}}
}

注：看清题目要求，输入的正整数是小于2^32，int类型的范围是-2^32~(2^31)-1,明显题目设坑了，输入的数范围超过了int类型，所以要用long类型来接收，至于求和sum的类型int,long都可以

方法二：利用递归求和

import java.util.Scanner;
import java.lang.Math;
public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);while(sc.hasNext()) {long a = sc.nextLong();long sum = 0;sum = s(a);System.out.println(sum);}}private static long s(long a) {if(a<10) {return a;}else {return a%10+s(a/10);}}
}

方法三：字符求和

import java.util.Scanner;
import java.lang.Math;
public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);while(sc.hasNext()) {String sts = sc.nextLine();int sum = 0;for(int i=0;i<sts.length();i++) {int a = sts.charAt(i) - '0';//减48也是可以的sum += a;}System.out.println(sum);}}
}