Java解题练习

Java解题练习

1.A+B问题

  解题代码：

import java.util.*;

public class Main {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		while(in.hasNext()) {//关于hasnext这个方法我可能是没有学过，这个方法有什么意义？
			int a = in.nextInt();
			int b = in.nextInt();
			System.out.println(a+b);
		}
	}
}

2.控制数量的A+B问题

  解题代码：

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int rownum = in.nextInt();
		int[] ret = new int[rownum];
		for(int i = 0;i<rownum;i++) {
			ret[i] = in.nextInt() + in.nextInt();
		}
		for(int i:ret) {
			System.out.println(i);
		}

	}

}

3.控制终止的A+B问题

  解题代码：

import java.util.Scanner;
import java.util.LinkedList;
public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		LinkedList<Integer> arr = new LinkedList();
		int a = 1,b = 1;
		while(true) {
			a = in.nextInt();
			b = in.nextInt();
			if(a==0&&b==0) {
				break;
			}
			arr.add(a+b);
		}

		for(int i:arr) {
			System.out.println(i);
		}

	}

}

4.一个自己做的

import java.util.Scanner;
import java.util.LinkedList;
public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		LinkedList<Integer> ls = new LinkedList();
		while(true) {
			int arr = 0;
			Scanner scanner = new Scanner(System.in);
			String s = scanner.nextLine();
			Scanner sc = new Scanner(s);
			while(sc.hasNext()) {
				arr+=sc.nextInt();
			}
			if(arr==0) {
				break;
			}
			ls.add(arr);
		}

		for(int i:ls) {
			System.out.println(i);
		}

	}

}

5.限制每行长度的多行A+B+C问题

  解题代码：

import java.util.Scanner;
import java.util.LinkedList;
public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner sc = new Scanner(System.in);
		LinkedList<Integer> ls = new LinkedList();
		while(true) {
			int leng = sc.nextInt();
			if(leng==0) {
				break;
			}
			int addr = 0;
			for(int i = 0;i<leng;i++) {
				addr+=sc.nextInt();
			}
			ls.add(addr);

		}

		for(int i:ls) {
			System.out.println(i);
		}

	}

}

6.有次数限制的A+B+C问题

  解题代码：

import java.util.Scanner;
import java.util.LinkedList;
public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner sc = new Scanner(System.in);
		LinkedList<Integer> ls = new LinkedList();
		int time = sc.nextInt();
		for(int j=0;j<time;j++) {
			int leng = sc.nextInt();
			int addr = 0;
			for(int i = 0;i<leng;i++) {
				addr+=sc.nextInt();
			}
			ls.add(addr);

		}

		for(int i:ls) {
			System.out.println(i);
		}

	}

}

7.我不太理解的hasnext版本的但是做出来的A+B+C问题

  解题代码：

import java.util.Scanner;
import java.util.LinkedList;
public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner sc = new Scanner(System.in);
		LinkedList<Integer> ls = new LinkedList();
		while(sc.hasNext()) {
			int leng = sc.nextInt();
			int addr = 0;
			for(int i = 0;i<leng;i++) {
				addr+=sc.nextInt();
			}
			System.out.println(addr);
		}
	}

}

8.每行后边接一个空行输出的A+B问题

  解题代码：

import java.util.Scanner;
import java.util.LinkedList;
public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()) {
			int sum = sc.nextInt()+sc.nextInt();
			System.out.println(sum);
			System.out.println();
		}

	}

}