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  • 1031. Hello World for U

    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  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 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
     1 #include<stdio.h>
     2 #include<math.h>
     3 #include<stdlib.h>
     4 #include<string.h>
     5 char a[100][100];
     6 int main()
     7 {
     8     char str[100];
     9     gets(str);
    10     int i = 0, j = 0, n1, n2, n = strlen(str);
    11     n1 = (n + 2) / 3;
    12     n2 = n + 2 - 2 * n1;
    13     for(i = 0; i < n1; i++)
    14         for(j = 0; j < n2; j++)
    15         {
    16             a[i][j] = ' ';
    17         }
    18     i = 0;
    19     for(j = 0; j < n1; j++)
    20     {
    21         a[j][0] = str[i++];
    22     }
    23     for(j = 1; j < n2; j++)
    24     {
    25         a[n1 - 1][j] = str[i++];
    26     }
    27     for(j = n1 - 2; j >= 0; j--)
    28     {
    29         a[j][n2 - 1] = str[i++];
    30     }
    31     for(i = 0; i < n1; i++)
    32     {
    33         for(j = 0; j < n2; j++)
    34         {
    35             printf("%c", a[i][j]);
    36         }
    37         printf("
    ");
    38     }
    39     return 0;
    40 }
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  • 原文地址:https://www.cnblogs.com/yomman/p/4270731.html
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