PAT (Advanced Level) Practice 1031 Hello World for U （20 分）

我不是女神ヾ 2022-05-07 01:00 213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

AC代码：

/*1031 Hello World for U
    给你一字符串，将它以'U型'输出 
    其中n1=n3分别表示U型的左右两侧的字符数，n2等于底部的字符数, N表示字符串里字符的总数 
    其中满足关系n1=n3=max{k|k<=n2, 3<=n2<=N}, n1+n2+n3-2=N 
    */
    #include <iostream>
    #include <string>
    using namespace std;
    int main(){
    	string s;
    	cin >> s;
    	int N = s.length();
    	int n1 = (N + 2) / 3;
    	int n2 =  N + 2 - 2*n1;
    	//将U型横着输出 
    	for(int i = 0; i < n1-1; i++) {
    		cout << s[i];
    		//输出U中间空的字符串 
    		for(int j = 0; j < n2-2; j++){
    			cout << " ";
    		}
    		cout << s[N-i-1] << endl;
    	}
    	//输出最底部的那一行字符串 
    	for(int k = 0; k < n2; k++) {
    		cout << s[n1-1+k];
    	}
    	return 0;
    }