【题意+推导讲解】1031 Hello World for U (20 分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

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 n1characters, 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

题意：输入一个字符串，按给定规则输出。

规则：

1、按样例中顺序输出字母
2、n1+n2+n3 = N+2
3、n2 >= 3
4、n1 <= n2

低级解法：定义双重循环，i循环为n2，从3开始遍历；j循环为n1，从i开始遍历，等于0时停止。找到解后，按给定规律输出。（解放双手，暴力万岁）

高级解法：手动推导，可得结论：n1=n3=(N+2)/3向下取整。n1,n2,n3确定以后按规定输出。

低级解法

#include<bits/stdc++.h>
using namespace std;
int main() {ios::sync_with_stdio(false);string s; cin >> s;int len = s.length(), b, l_r;for(int i = 3; ; i++)                     //求n1和n2 for(int j = i; j >= 1; j--) if(j*2 + i == len+2) {b = i; l_r = j-1; goto loop;}loop:;//输出 int k = 0;   //计数器 for(int i = 0; i < l_r+1; i++) {   //行数 for(int j = 0; j < b; j++) { //列数 if(i != l_r) if(j == 0) putchar(s[k]);else if(j == b-1) putchar(s[len- k++ -1]);  else putchar(' ');else putchar(s[k++]); } putchar('\n'); }return 0;
}

高级解法

#include<bits/stdc++.h>
using namespace std;
int main(){string input;  cin>> input//读取字符串int n1=(input.size()+2)/3,n2=input.size()+2-2*n1;//获取n1,n2//进行输出for(int i=0;i<n1-1;++i){//前n1-1行，需要输出首尾两个字符，且每行字符数为n2printf("%c",input[i]);for(int j=0;j<n2-2;++j)printf(" ");printf("%c\n",input[input.size()-i-1]);}//最后一行将没有输出的n2个字符一次性输出for(int i=0;i<n2;++i)printf("%c",input[n1-1+i]);return 0;
}

耗时：

【题意+推导讲解】1031 Hello World for U (20 分)_15行代码AC相关推荐

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【题意+推导讲解】1031 Hello World for U (20 分)_15行代码AC

立志用最少的代码做最高效的表达

低级解法

高级解法

耗时：

【题意+推导讲解】1031 Hello World for U (20 分)_15行代码AC相关推荐

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