CPE练习(一):11764 Cricket Field——JAVA

亦凉 2022-04-04 14:46 77阅读 0赞

### CPE题目 ###

题目链接:  
[https://uva.onlinejudge.org/external/117/p11764.pdf][https_uva.onlinejudge.org_external_117_p11764.pdf]  
题目文字:  
Mario is in the final castle. He now needs to  
jump over few walls and then enter the Koopa’s  
Chamber where he has to defeat the monster in  
order to save the princess. For this problem,  
we are only concerned with the “jumping over  
the wall” part. You will be given the heights  
of N walls from left to right. Mario is currently  
standing on the first wall. He has to jump to the  
adjacent walls one after another until he reaches  
the last one. That means, he will make (N − 1)  
jumps. A high jump is one where Mario has to  
jump to a taller wall, and similarly, a low jump is one where Mario has to jump to a shorter wall. Can you find out the total number of high jumps and low jumps Mario has to make?

*  Input  
    The first line of input is an integer T (T < 30) that indicates the number of test cases. Each case starts with an integer N (0 < N < 50) that determines the number of walls. The next line gives the height of the N walls from left to right. Each height is a positive integer not exceeding 10.
 *  Output  
    For each case, output the case number followed by 2 integers, total high jumps and total low jumps, respectively. Look at the sample for exact format.
 *  Sample Input  
    3  
    8 14223534 1  
    9  
    5  
    12345
 *  Sample Output  
    Case 1: 4 2  
    Case 2: 0 0  
    Case 3: 4 0

代码如下:

import java.util.Scanner;
    
    public class lyx {
    
    	public static void main(String[] args) {
    		// TODO Auto-generated method stub
    		Scanner scanner=new Scanner(System.in);
    		int number=scanner.nextInt();
    		for(int num=1;num<=number;num++)
    		{
    		int count=scanner.nextInt();//每次输入数字
    		int [] data=new int[50];//创建data数组 
    			 for(int i=0;i<count;i++)	
    			 {	 data[i]=scanner.nextInt();	 }
    			 int max=0 , min=0;
    			 for(int i=0;i<count-1;i++)	
    			 {	
    				 if((data[i+1]-data[i])>0)
    				 {max++;}
    				 else if((data[i+1]-data[i])<0)
    				 {min++;}
    			 }
    			 System.out.println("Case "+num+": "+max+" "+min);
    
    		}
    
    	}
    
    }

[https_uva.onlinejudge.org_external_117_p11764.pdf]: https://uva.onlinejudge.org/external/117/p11764.pdf