java刷题--蓝桥杯第九届JavaC组1--3

蔚落 2024-03-26 14:45 68阅读 0赞

原题链接：[https://blog.csdn.net/feng8403000/article/details/123303285][https_blog.csdn.net_feng8403000_article_details_123303285]

1.  ## 哪天返回 ##

### 题目 ###

小明被不明势力劫持。后被扔到x星站再无问津。小明得知每天都有飞船飞往地球，但需要108元的船票，而他却身无分文。

他决定在x星战打工。好心的老板答应包食宿，第1天给他1元钱。

并且，以后的每一天都比前一天多2元钱，直到他有足够的钱买票。

请计算一下，小明在第几天就能凑够108元，返回地球

### 问题分析 ###

1.  通过计数累加计算钱数

1.  通过累加计算每日工资

1.  累加或者函数循环计数天数

1.  注意是第二天加工资所有要先计算有多少钱，累加之后，在增加工资

### 代码实现 ###

package 省赛一轮;
    
    public class 第9届题目1 {
    
        public static void main(String[] args) {
            // TODO Auto-generated method stub
    /*
    但需要108元的船票，
     * 而他却身无分文。
    第1天给他1元钱。
    并且，以后的每一天都比前一天多2元钱，直到他有足够的钱买票。
    请计算一下，小明在第几天就能凑够108元，返回地球。
    */
            int a = 1;
            int sum = 0;
            int day = 0;
            for (int i = 0; sum < 108; i++) {
                sum+=a;
                a+=2;            
                day++;
            }
            System.out.println(day);
        }
    
    }

输出 11

## 2、猴子分香蕉 ##

### 题目 ###

5只猴子是好朋友，在海边的椰子树上睡着了。这期间，有商船把一大堆香蕉忘记在沙滩上离去。

第1只猴子醒来，把香蕉均分成5堆，还剩下1个，就吃掉并把自己的一份藏起来继续睡觉。

第2只猴子醒来，重新把香蕉均分成5堆，还剩下2个，就吃掉并把自己的一份藏起来继续睡觉。

第3只猴子醒来，重新把香蕉均分成5堆，还剩下3个，就吃掉并把自己的一份藏起来继续睡觉。

第4只猴子醒来，重新把香蕉均分成5堆，还剩下4个，就吃掉并把自己的一份藏起来继续睡觉。

第5只猴子醒来，重新把香蕉均分成5堆，哈哈，正好不剩！

请计算一开始最少有多少个香蕉。

### 问题分析 ###

总数经过变化满足所有条件

每次余数满足下一个条件，也就是连环&&

五个条件同时满足就是符合条件的值

### 代码实现 ###

package 省赛一轮;
    
    public class 第9届题目21 {
    
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            
            for (int num = 0; num < 10000; num++) {    
                int n;
                if (num%5==1) {
                    n = num-(num/5)+1;
                    if (n%5==2) {
                        n = n-(n/5)+2;
                        if (n%5==3) {
                            n = n-(n/5)+3;
                            if (n%5==4) {
                                n = n-(n/5)+4;
                                if (n%5==0) {
                                    System.out.println(num);
                                }
                            }
                        }
                    }
                }
            }
        }
    
    }

## 3、字母阵列 ##

### 题目 ###

仔细寻找，会发现：在下面的8x8的方阵中，隐藏着字母序列：“LANQIAO”。

SLANQIAO

ZOEXCCGB

MOAYWKHI

BCCIPLJQ

SLANQIAO

RSFWFNYA

XIFZVWAL

COAIQNAL

我们约定: 序列可以水平，垂直，或者是斜向；

并且走向不限（实际上就是有一共8种方向）。

上图中一共有4个满足要求的串。

下面有一个更大的（100x100）的字母方阵。

你能算出其中隐藏了多少个“LANQIAO”吗？

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    NMHGYPFUUDWKQGTAKKGCDFJFYJFNRZVXDPGZEAMWQVQZODKTXHIYFVKJSSAWVHYCUCZMLLBPXTILDYJQEMWDRUFKISOUVPUDTYPB
    FDAQUBXHUJYTAYNWVIJNUSQDTQDEMUAPWXRYUWONTBDZCHZOUEGPMWEZTQWWSHAYOBWVTDIMZYNVNZKUHOFCQKPHJXWNRCGUJEKO
    WSDAUGUTVWCVHEMOIRJJGTANUWTSAIXXEVZTBDHPGSRHHVWCDZVZYRJTLONIJVXEATHQXOUKBIGZONFRSZIOGWNTYAJYLQCGEOWY

### 问题分析 ###

遇到 L 就搜索周边是否有 A

有 --》 搜索 N

无--》 结束

类推

直到O 计数加1，结束循环

搜索方式

在二维数组里

向八个方向搜索，arr i+-1 j+-1

双for遍历不成立，，inx与iny对应组成八种移动

### 代码实现 ###

package 省赛一轮;
    
    import java.util.Scanner;
    
    public class 第9届题目3 {
        static int count = 0;
        static int[] inx = {0,-1,1,1,0,-1,-1,-1};
        static int[] iny = {1,1,0,1,-1,1,0,-1};
        static char[] target = {'L','A','N','Q','I','A','O'};
        static char[][] arr = new char[100][100];
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            Scanner sc = new Scanner(System.in);
    
    
            for (int i = 0; i < arr.length; i++) {
                arr[i] = sc.nextLine().toCharArray();
    
            }
            for (int i = 0; i < arr.length; i++) {
                for (int j = 0; j < arr.length; j++) {
                    if (arr[i][j]==target[0]) {
                        arrmove(i,j);
                    }
                }
            }
            System.out.println(count);
            sc.close();
        }
    
        private static void arrmove(int i, int j) {
    
            for (int s = 0; s < 8; s++) {
                int x = i + inx[s];
                int y = j + iny[s];
                int n = 0;
                while (x<100&&x>=0&&y<100&&y>=0&&arr[x][y]==target[n]) {
                    if (n == 6) {
                        count++;
                        break;
                    }
                    n++; 
                
                    x += inx[s];
                    y += iny[s];
    
                }
            }
        }
    
    }

输出结果 1？

[https_blog.csdn.net_feng8403000_article_details_123303285]: https://blog.csdn.net/feng8403000/article/details/123303285