PAT/图形输出习题集

B1027. 打印沙漏 (20)

Description：

本题要求你写个程序把给定的符号打印成沙漏的形状。例如给定17个“*”，要求按下列格式打印

*****
 ***
  *
 ***
*****

所谓“沙漏形状”，是指每行输出奇数个符号；各行符号中心对齐；相邻两行符号数差2；符号数先从大到小顺序递减到1，再从小到大顺序递增；首尾符号数相等。

给定任意N个符号，不一定能正好组成一个沙漏。要求打印出的沙漏能用掉尽可能多的符号。

Input：

输入在一行给出1个正整数N（<=1000）和一个符号，中间以空格分隔。

Output：

首先打印出由给定符号组成的最大的沙漏形状，最后在一行中输出剩下没用掉的符号数。

Sample Input：

19 *

Sample Output：

*****
  ***
   *
  ***
*****
2

 #include <cstdio>
 #include <cmath>

 int main()
 {
     int n;
     char c;
     scanf("%d %c", &n, &c);

     int bottom = (int)sqrt(2.0*(n+))-;
     if(bottom% == )
         --bottom;
     int used = (bottom+)*(bottom+)/-;
     for(int i=bottom; i>=; i-=) {
         for(int j=; j<(bottom-i)/; ++j)   printf(" ");
         for(int j=; j<i; ++j)  printf("%c", c);
         printf("\n");
     }
     for(int i=; i<=bottom; i+=) {
         for(int j=; j<(bottom-i)/; ++j)   printf(" ");
         for(int j=; j<i; ++j)  printf("%c", c);
         printf("\n");
     }
     printf("%d\n", n-used);

     return ;
 }

A1031. Hello World for U (20)

Description：

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 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：

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：

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

 #include <cstdio>
 #include <cstring>

 int main()
 {
     char str[], ans[][];
     gets(str);

     int N = strlen(str);
     int n1 = (N+)/, n3 = n1, n2 = N+-n1-n3;
     for(int i=; i<=n1; ++i) {
         for(int j=; j<=n2; ++j)
             ans[i][j] = ' ';
     }

     int pos = ;
     for(int i=; i<=n1; ++i)    ans[i][] = str[pos++];
     for(int j=; j<=n2; ++j)    ans[n1][j] = str[pos++];
     for(int i=n3-; i>=; --i)  ans[i][n2] = str[pos++];
     for(int i=; i<=n1; ++i) {
         for(int j=; j<=n2; ++j)
             printf("%c", ans[i][j]);
         printf("\n");
     }

     return ;
 }

 #include <cstdio>
 #include <cstring>

 int main()
 {
     char str[];
     gets(str);

     int N = strlen(str);
     int n1 = (N+)/, n3 = n1, n2 = N+-n1-n3;
     for(int i=; i<n1-; ++i) {
         printf("%c", str[i]);
         for(int j=; j<n2-; ++j)
             printf(" ");
         printf("%c\n", str[N-i-]);
     }
     for(int i=; i<n2; ++i)
         printf("%c", str[n1+i-]);

     return ;
 }