1031 Hello World for U (20 分)
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 n2characters, 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
题意:
将给定的字符串照U型输出。
思路:
找到小于len+2的最大的三的倍数,(len+2)/3就是竖着的个数。
1 #include <iostream> 2 #include <set> 3 #include <cmath> 4 #include <stdio.h> 5 #include <cstring> 6 #include <algorithm> 7 #include <vector> 8 #include <queue> 9 #include <map> 10 using namespace std; 11 typedef long long LL; 12 #define inf 0x7f7f7f7f 13 14 char s[90]; 15 16 int main() 17 { 18 scanf("%s", s); 19 int n = strlen(s); 20 int x = (n + 2) / 3; 21 int y = n - 2 * x; 22 23 for(int i = 0; i < x - 1; i++){ 24 printf("%c", s[i]); 25 for(int j = 0; j < y; j++){ 26 printf(" "); 27 } 28 printf("%c ", s[n - 1 - i]); 29 } 30 for(int i = x - 1; i <= x + y; i++){ 31 printf("%c", s[i]); 32 } 33 printf(" "); 34 35 return 0; 36 }