1031. Hello World for U (20)

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1031. Hello World for U (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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  de  ll  rlowo

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   dl   llowor
这题很简单，解题关键点，分析清楚n1,n2的值怎么表达。由题由n1<=n2 && n1*2 +n2+2= n分析可得n1 = (s.length()+2) / 3，这里我一开始都没分析清楚，也是服了自己了，这里确实不难，正式考试的时候，更不应该浪费掉更多的时间
#include<vector>#include <sstream>#include<cmath>#include<iomanip>#include<iostream>#include <ctype.h>#include <stdlib.h>#include <algorithm>using namespace std;int main(){string s;cin >> s;int n1 = (s.length()+2) / 3;//由n1<=n2 && n1*2 +n2+2= n分析可得，这里我一开始都没分析清楚int n2 = s.length() - 2 * n1 + 2;for (int i = 0; i < n1 - 1; i++){cout << s[i];for (int j = 0; j < n2-2; j++){cout << " ";}cout << s[s.length() - 1 - i];cout << endl;}for (int i = n1 - 1; i < n1 - 1 + n2; i++){cout << s[i];}return 0;}

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