B1027. 打印沙漏 (20)
Description：
本題要求你寫個程序把給定的符號打印成沙漏的形狀。例如給定17個“*”，要求按下列格式打印
***** *** * *** *****
所謂“沙漏形狀”，是指每行輸出奇數個符號；各行符號中心對齊；相鄰兩行符號數差2；符號數先從大到小順序遞減到1，再從小到大順序遞增；首尾符號數相等。
給定任意N個符號，不一定能正好組成一個沙漏。要求打印出的沙漏能用掉盡可能多的符號。
Input：
輸入在一行給出1個正整數N（<=1000）和一個符號，中間以空格分隔。
Output：
首先打印出由給定符號組成的最大的沙漏形狀，最後在一行中輸出剩下沒用掉的符號數。
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 }
