Codeforces Round #259 (Div. 2)AB

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

A. Little Pony and Crystal Mine

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

3

output

*D*
DDD
*D*

input

5

output

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

input

7

output

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

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

 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>

 int main()
 {
     int i,j,n,m,k,t;
     while(scanf("%d",&n)!=EOF)
     {
         int tmp1=n/,tmp2=n/,tmp3=;
         for(i=;i<=n/;i++)
         {
             for(j=;j<=tmp1;j++)
                 printf("*");
             for(j=tmp1;j<=tmp2;j++)
                 printf("D");
             for(j=tmp2+;j<=n;j++)
                 printf("*");
             printf("\n");
             tmp1--;tmp2++;
         }
         for(i=;i<=n;i++)
             printf("D");
         printf("\n");
         tmp1=,tmp2=n;
         for(i=;i<=n/;i++)
         {
             for(j=;j<=tmp1;j++)
                 printf("*");
             for(j=tmp1+;j<=tmp2-;j++)
                 printf("D");
             for(j=tmp2;j<=n;j++)
                 printf("*");
             tmp1++;tmp2--;
             printf("\n");
         }
     }
     return ;
 }

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

B. Little Pony and Sort by Shift

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

1

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

 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>

 int str[];

 int main()
 {
     int n,m,i,j;
     while(scanf("%d",&n)!=EOF)
     {
         for(i=; i<=n; i++)
         {
             scanf("%d",&str[i]);
         }
         int sum=,tmp;
         for(i=; i<=n; i++)
         {
             if(str[i]<str[i-])
             {
                 sum++;
                 tmp=i;
             }

         }
         if(sum == )printf("0\n");
         else if(sum > )printf("-1\n");
         else if(sum ==  && str[n] <= str[])
             printf("%d\n",n-tmp+);
         else printf("-1\n");
     }
     return ;
 }

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……