1031 Hello World for U (20 point(s)) - C語言 PAT 甲級

1031 Hello World for U (20 point(s))

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   !
e   d
l    l
lowor

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

題目大意：

輸入的字符串，按照 U 字型輸出；

n₁ 和 n₃ 表示兩條豎線上字符的總個數，n₂ 表示底部橫線上字符的總個數，並要求：

1. n₁ == n₃
2. n₂ >= n₁
3. n₁ 取儘可能大的值

設計思路：

難點在於確定 n₁、n₂ 的值；

因爲 U 分爲 n₁、n₂、n₃ 三條邊，且重疊了兩個字符，所以設：

• n = 原字符串長度 + 2 = n₁ * 2 + n₂；
  • 若 n % 3 == 0，則 n₁ == n₂ == n₃；
  • 若 n % 3 > 0，並且需要 n₂ >= n₁，那麼把多出的餘數給 n₂ 就行了；
• 所以 n₁ = n / 3，n₂ = n / 3 + n % 3；

確定了三條邊的長度，利用一個頭指針和一個尾指針，按行輸出原字符串即可

編譯器：C (gcc)

#include <stdio.h>
#include <string.h>

int main(void)
{
        char str[81], *p, *q;
        int n1, n2, count;
        int i, j;

        scanf("%s", str);

        count = strlen(str) + 2;
        p = str;
        q = str + (count - 2 - 1);
        n1 = count / 3;
        n2 = count / 3 + count % 3;
        for (i = 0; i < n1 - 1; i++) {
                printf("%c", *p);
                p++;
                for (j = 1; j < n2 - 1; j++)
                        printf(" ");
                printf("%c\n", *q);
                q--;
        }
        for (p; p <= q; p++)
                printf("%c", *p);

        return 0;
}