51nod_NOIP普及组预科班模拟赛Java题解

本文提供了NOIP普及组模拟赛中五个编程题目的解决方案,包括基本的数值计算、条件判断、溢出检查、数组操作及累积求和等核心算法。每个题目都附带了完整的Java代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

A:

 题解:

package a51nodWeb.NOIP普及组模拟赛;

import java.util.Scanner;

public class A {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int K = in.nextInt();
		int sum = 0;
		int j=1;
		int k=j;
		for (int i = 0; i < K; i++) {
			sum+=k;
			j--;
			if(j==0) {
				k=k+1;
				j=k;
			}
		}
		System.out.println(sum);
	}

}

B:

 

package a51nodWeb.NOIP普及组模拟赛;

import java.util.Scanner;

public class B {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		if(n%2==0) {
			System.out.println("0");
		}else {
			System.out.println("1");
		}
	}

}

 C:

 

package a51nodWeb.NOIP普及组模拟赛;

import java.util.Scanner;

public class C {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		long a = in.nextLong();
		long b = in.nextLong();
		if(a+b>Integer.MAX_VALUE || a+b<Integer.MIN_VALUE) {
			System.out.println("YES");
		}else {
			System.out.println("NO");
		}
	}

}

 D:

package a51nodWeb.NOIP普及组模拟赛;

import java.util.Scanner;

public class D {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] nums = new int[n];
		for (int i = 0; i < nums.length; i++) {
			nums[i] = in.nextInt();
		}
		System.out.println(n);
		for (int i = nums.length-1; i >=0; i--) {
			System.out.println(nums[i]);
		}
	}

}

E:

package a51nodWeb.NOIP普及组模拟赛;

import java.util.Scanner;

public class E {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] nums = new int[n];
		long[] ak = new long[n];
		nums[0] = in.nextInt();
		ak[0] = nums[0];
		for (int i = 1; i < nums.length; i++) {
			nums[i] = in.nextInt();
			ak[i]+=ak[i-1]+nums[i];
		}
		System.out.println(n);
		for (int i = 0; i < ak.length; i++) {
			System.out.println(ak[i]);
		}
	}

}

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值