Codeforces Round #259 (Div. 2)AB

链接：http://codeforces.com/contest/454/problem/A

Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n × n matrix with a diamond inscribed into it.

You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.

Input

The only line contains an integer n (3 ≤ n ≤ 101; n is odd).

Output

Output a crystal of size n.

Sample test(s)

input

output

*D*
DDD
*D*

input

output

**D**
*DDD*
DDDDD
*DDD*
**D**

input

output

***D***
**DDD**
*DDDDD*
DDDDDDD
*DDDDD*
**DDD**
***D***


。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。
。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。
看样例就知道了，直接模拟即可

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <stdlib.h>
 4 
 5 int main()
 6 {
 7     int i,j,n,m,k,t;
 8     while(scanf("%d",&n)!=EOF)
 9     {
10         int tmp1=n/2,tmp2=n/2,tmp3=1;
11         for(i=1;i<=n/2;i++)
12         {
13             for(j=1;j<=tmp1;j++)
14                 printf("*");
15             for(j=tmp1;j<=tmp2;j++)
16                 printf("D");
17             for(j=tmp2+2;j<=n;j++)
18                 printf("*");
19             printf("
");
20             tmp1--;tmp2++;
21         }
22         for(i=1;i<=n;i++)
23             printf("D");
24         printf("
");
25         tmp1=1,tmp2=n;
26         for(i=1;i<=n/2;i++)
27         {
28             for(j=1;j<=tmp1;j++)
29                 printf("*");
30             for(j=tmp1+1;j<=tmp2-1;j++)
31                 printf("D");
32             for(j=tmp2;j<=n;j++)
33                 printf("*");
34             tmp1++;tmp2--;
35             printf("
");
36         }
37     }
38     return 0;
39 }

B：http://codeforces.com/contest/454/problem/B

One day, Twilight Sparkle is interested in how to sort a sequence of integers a1, a2, ..., an in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:

a1, a2, ..., an → an, a1, a2, ..., an - 1.

Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?

Input

The first line contains an integer n (2 ≤ n ≤ 105). The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.

Sample test(s)

input

2
2 1

output

input

3
1 3 2

output

-1

input

2
1 2

output

0

，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，
//////////////////////////////////////////////////////
模拟题，判断每两个出现递减的，算出有多少个递减的，如果有两个以上的，就是不行的，
当为零，说明没有递减的，为一需判断起点与终点的大小，可不可以接上

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <stdlib.h>
 4 
 5 int str[100010];
 6 
 7 int main()
 8 {
 9     int n,m,i,j;
10     while(scanf("%d",&n)!=EOF)
11     {
12         for(i=1; i<=n; i++)
13         {
14             scanf("%d",&str[i]);
15         }
16         int sum=0,tmp;
17         for(i=2; i<=n; i++)
18         {
19             if(str[i]<str[i-1])
20             {
21                 sum++;
22                 tmp=i;
23             }
24                 
25         }
26         if(sum == 0)printf("0
");
27         else if(sum > 1)printf("-1
");
28         else if(sum == 1 && str[n] <= str[1])
29             printf("%d
",n-tmp+1);
30         else printf("-1
");
31     }
32     return 0;
33 }

今天没事干，模拟了下CF，只做了A题，模拟速度超慢　　A题模拟了15min，B题想了一会没思路，就取看C题，找了一会规律没找到，

就这样结束了……

C题组合数学，看不太懂题解，
AC不易，且行且珍惜……