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PAT/图形输出习题集

2018-06-18 04:13:07来源：未知阅读 ()

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

 1 #include <cstdio>
 2 #include <cmath>
 3 
 4 int main()
 5 {
 6     int n;
 7     char c;
 8     scanf("%d %c", &n, &c);
 9 
10     int bottom = (int)sqrt(2.0*(n+1))-1;
11     if(bottom%2 == 0)
12         --bottom;
13     int used = (bottom+1)*(bottom+1)/2-1;
14     for(int i=bottom; i>=1; i-=2) {
15         for(int j=0; j<(bottom-i)/2; ++j)   printf(" ");
16         for(int j=0; j<i; ++j)  printf("%c", c);
17         printf("\n");
18     }
19     for(int i=3; i<=bottom; i+=2) {
20         for(int j=0; j<(bottom-i)/2; ++j)   printf(" ");
21         for(int j=0; j<i; ++j)  printf("%c", c);
22         printf("\n");
23     }
24     printf("%d\n", n-used);
25 
26     return 0;
27 }

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

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 int main()
 5 {
 6     char str[100], ans[40][40];
 7     gets(str);
 8 
 9     int N = strlen(str);
10     int n1 = (N+2)/3, n3 = n1, n2 = N+2-n1-n3;
11     for(int i=1; i<=n1; ++i) {
12         for(int j=1; j<=n2; ++j)
13             ans[i][j] = ' ';
14     }
15 
16     int pos = 0;
17     for(int i=1; i<=n1; ++i)    ans[i][1] = str[pos++];
18     for(int j=2; j<=n2; ++j)    ans[n1][j] = str[pos++];
19     for(int i=n3-1; i>=1; --i)  ans[i][n2] = str[pos++];
20     for(int i=1; i<=n1; ++i) {
21         for(int j=1; j<=n2; ++j)
22             printf("%c", ans[i][j]);
23         printf("\n");
24     }
25 
26     return 0;
27 }

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 int main()
 5 {
 6     char str[100];
 7     gets(str);
 8 
 9     int N = strlen(str);
10     int n1 = (N+2)/3, n3 = n1, n2 = N+2-n1-n3;
11     for(int i=0; i<n1-1; ++i) {
12         printf("%c", str[i]);
13         for(int j=0; j<n2-2; ++j)
14             printf(" ");
15         printf("%c\n", str[N-i-1]);
16     }
17     for(int i=0; i<n2; ++i)
18         printf("%c", str[n1+i-1]);
19 
20     return 0;
21 }