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。
#include <iostream>
#include <set>
#include <cmath>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
using namespace std;
typedef long long LL;
#define inf 0x7f7f7f7f char s[]; int main()
{
scanf("%s", s);
int n = strlen(s);
int x = (n + ) / ;
int y = n - * x; for(int i = ; i < x - ; i++){
printf("%c", s[i]);
for(int j = ; j < y; j++){
printf(" ");
}
printf("%c\n", s[n - - i]);
}
for(int i = x - ; i <= x + y; i++){
printf("%c", s[i]);
}
printf("\n"); return ;
}