1. 题目
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 n_1 characters, then left to right along the bottom line with _n_2 characters, and finally bottom-up along the vertical line with _n_3 characters. And more, we would like U to be as squared as possible — that is, it must be satisfied that _n_1= _n_3= _max { k | k≤ n_2 for all 3≤ _n_2≤ _N } with n_1+ _n_2+ _n_3−2= _N.
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.
For each test case, print the input string in the shape of U as specified in the description.
helloworld!
h !
e d
l l
lowor
2. 题意
将一串字符串以 U型形式输出出来。
3. 思路——字符串
根据题意计算n1,n2,n3，这里n1等于n3，只要定义n1和n2即可。
计算方法：
已知：n1=n3=max {k | k≤n2 for all 3≤n2≤N }，且n1+n2+n3-2=N
可得：(n1=n3=(N+2)/2)（结果向下取整）
(n2=N+2-n1-n3)
计算出n1和n2后，即可根据题目要求输出 U型图形（见代码）。
4. 代码
#include
#include
using namespace std;
int main()
{
string str;
cin >> str;
int n1, n2;
int N = str.length();
n1 = (N + 2) / 3;
n2 = N + 2 - (2 * n1);
for (int i = 0; i < n1 - 1; ++i)
{
// 输出第i个字符
cout << str[i];
// 输出中间的空格
for (int j = 0; j < n2 - 2; ++j) cout << " ";
// 输出倒数第i+1个字符
cout << str[N - 1 - i] << endl;
}
// 最后一行输出剩下的中间字符串
cout << str.substr(n1 - 1, n2) << endl;
return 0;
}
Original: https://www.cnblogs.com/vanishzeng/p/15484819.html
Author: vanish丶
Title: 1031 Hello World for U (20 分)
原创文章受到原创版权保护。转载请注明出处：https://www.johngo689.com/584670/
转载文章受原作者版权保护。转载请注明原作者出处！