PAT甲1031 Hello World for U【字符串】

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 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

 

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[90];

int main()
{
    scanf("%s", s);
    int n = strlen(s);
    int x = (n + 2) / 3;
    int y = n - 2 * x;

    for(int i = 0; i < x - 1; i++){
        printf("%c", s[i]);
        for(int j = 0; j < y; j++){
            printf(" ");
        }
        printf("%c\n", s[n - 1 - i]);
    }
    for(int i = x - 1; i <= x + y; i++){
        printf("%c", s[i]);
    }
    printf("\n");

    return 0;
}