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("
");
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("
");
23 }
24 printf("%d
", 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("
");
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
", 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 }