1031 Hello World for U (20 point(s))

題目原文

??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.

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

題目大意

??題目給出一字符串，然后要求把它打印成U形狀

解題思路

??水題，不解釋

代碼

#include <iostream> #include <string> using namespace std;int main() {string str;cin >> str;int n1, n2;for (n2 = 3; n2 < str.size(); ++n2) {for (n1 = n2; n1 > 0; --n1) {if (2 * n1 + n2 - 2 == str.size()) {for (int i = 0; i < n1 - 1; ++i) {cout << str[i] << string(n2 - 2, ' ') << str[str.size() - i - 1] << endl;}cout << str.substr(n1 - 1, n2) << endl;return 0;}else if (2 * n1 + n2 - 2 < str.size()) break;}}return 0; }

總結

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