PAT甲级真题及训练集(11)--1031. Hello World for U (20)

1031. Hello World for U (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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  de  ll  rlowo

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n₁characters, 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   dl   llowor

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提交代码

/**作者：一叶扁舟时间：17:10 2017/6/25思路：*/#include <stdio.h>#include <stdlib.h>#include <string.h>#define SIZE 10001int main(){int n1, n2, n3;int N;//总共字符总数char cha[SIZE];scanf("%s",cha);//计算输入的字符总长度N = strlen(cha);//printf("长度:%d", N);n1 = (N + 2) / 3 - 1;//两侧的数量，不包含最后一行的字母n3 = n1;n2 = N - n1 - n3;//最后一行的长度char left[SIZE];char middle[SIZE];char right[SIZE];strncpy(left, cha, n1);strncpy(middle, cha + n1, n2);strncpy(right, cha + n1 + n2, n3);middle[n2] = '\0';for (int i = 0; i < n1; i++){//输出最左边的字符printf("%c", left[i]);//输出中间的空格for (int j = 0; j < n2 - 2; j++)printf(" ");//输出最右边的字符,倒着输出int k = n3 - i - 1;printf("%c\n", right[k]);}//最后一行printf("%s\n", middle);system("pause");return 0;}