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

思路：n2 >= (N/3)+ 2/3
n2 >= 3；
邊界為7
另一種做法：(n+2)%3=0 : n1=n2=n3
(n+2)%3 !=0 : n1 = n3 =(n+2)/3

#include <iostream> #include <cmath>using namespace std;int main() {string s;cin >> s;int len = s.length();if(len<7){printf("%c", s[0]);for(int i=0;i<len-2-2;i++)printf(" ");printf("%c\n", s[len-1]);for(int i=1;i<len-1;i++){printf("%c", s[i]);}}else{int n2=ceil(len/3.0+2/3.0);if((len-n2)%2 != 0)n2 = n2+1;for(int i=0;i<(len-n2)/2;i++){printf("%c", s[i]);for(int j=0;j<n2-2;j++){printf(" ");}printf("%c\n", s[len-1-i]);}for(int i=(len-n2)/2;i<(len-n2)/2+n2;i++){printf("%c", s[i]);}}return 0; }

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