Given any string of N (≥) 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 } 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
n1+n2+n3-2 = n;
n1=n3所以-> 2*n1 + n2 = n+2
因为n2>n1
所以n1 = n/3, n2 = n/3 + n%3
#include<iostream> #include<string.h> using namespace std; int main() { char c[81]={' '}; char u[30][30]; scanf("%s",&c); memset(u,' ',sizeof(u)); int n = strlen(c)+2; int n1 = n/3; int n2 = n/3 + n%3; int j=0; for(int i=0;i<n1;i++) u[i][0] = c[j++]; for(int i=1;i<=n2-2;i++)//注意要减去二,因为倒u型的底部有两个重合 u[n1-1][i] = c[j++]; for(int i=n1-1;i>=0;i--) u[i][n2-1]=c[j++]; for(int i=0;i<n1;i++) { for(int j=0;j<n2;j++) { printf("%c",u[i][j]); } printf(" "); } return 0; }