PAT 甲级图形输出 1031 Hello World for U (20 分)

短命女 2024-02-19 20:55 43阅读 0赞

1031 Hello World for U (20 分)

Given any string of N (≥5) characters, you are asked to form the characters into the shape of `U`. For example, `helloworld` can be printed as:

h  d
    e  l
    l  r
    lowo

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like `U` to be as squared as possible -- that is, it must be satisfied that n1=n3=max \{ k | k≤n2 for all 3≤n2≤N \} with n1+n2+n3−2=N.

#### Input Specification: ####

Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

#### Output Specification: ####

For each test case, print the input string in the shape of U as specified in the description.

#### Sample Input: ####

helloworld!

#### Sample Output: ####

h   !
    e   d
    l   l
    lowor

我的思路：

取字符串的总长度，看是否能整除3，（令整除3的商为a，余数为b）

如果能，n1部分长度取a，n2部分长度取a+2，n3部分长度为a，

如果不能，n1部分长度取a+1，n2部分长度取a+b，n3部分长度为a+1

具体看代码：

#include<iostream>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    int main()
    {
    	string s;
    	int len=0;
    	int i,j,k,t;
    	cin>>s;
    	len=s.length();
    	if(len%3==0)
    	{
    		k=len/3;
    		j=0;
    		for(i=0;i<k;i++)
    		{
    			if(i!=k-1)
    			{
    				printf("%c",s[j]);
    				for(t=0;t<k;t++)
    				{
    					printf(" ");
    				}
    				printf("%c",s[len-j-1]);
    			}
    			else
    			{
    				for(t=0;t<k+2;t++)
    				printf("%c",s[t+j]);
    			}
    			j++;
    			printf("\n");
    		}
    	}
    	else
    	{
    		k=len/3;
    		j=0;
    		for(i=0;i<k;i++)
    		{
    			printf("%c",s[j]);
    			for(t=0;t<k+len%3-2;t++)
    			{
    				printf(" ");
    			}
    			printf("%c",s[len-j-1]);
    			j++;
    			printf("\n");
    		}
    		for(i=0;i<k+len%3;i++)
    		{
    			printf("%c",s[j+i]);
    		}
    		printf("\n");
    	}
    	return 0;
    }

## 别人的代码： ##

我每次几乎每次都找一个比自己简洁的代码

但我真的佩服这位博主的代码

方法：求出字符串长度 n , 然后得 m = n+2  
然后再得到 n1=n3=m/3，同时n2 = m - n1\*2；  
我写的代码中求出的n2是比上面的n2少2.

#include <bits/stdc++.h> 
    using namespace std;
    int main() {
    	//freopen("datain.txt","r",stdin); 
    	string s;
    	cin>>s;
    	int n = s.length();
    	//cout<<n<<endl;
    	int n1 = (n+2)/3;
    	int n2 = n-n1*2;
    	//cout<<n1<<" "<<n2<<endl;
    	string s1 = s.substr(0,n1);//n1字符串
    	string s2 = s.substr(n1,n2);//n2字符串
    	string s3 = s.substr(n1+n2);//n3字符串
    	//cout<<s1<<" "<<s2<<" "<<s3<<endl;
    	for (int i = 0; i < n1; i++) {
    		cout<<s1[i];
    		for (int j = 0; j < n2; j++) {
    			if (i != n1-1) //中间空格 
    				cout<<" ";
    			else //最后一行
    				cout<<s2[j];
    		}
    		cout<<s3[n1-i-1]<<endl;
    	}
    	return 0;
    }

加油！

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