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/2,tmp2=n/2,tmp3=1;
        for(i=1;i<=n/2;i++)
        {
            for(j=1;j<=tmp1;j++)
                printf("*");
            for(j=tmp1;j<=tmp2;j++)
                printf("D");
            for(j=tmp2+2;j<=n;j++)
                printf("*");
            printf("
");
            tmp1--;tmp2++;
        }
        for(i=1;i<=n;i++)
            printf("D");
        printf("
");
        tmp1=1,tmp2=n;
        for(i=1;i<=n/2;i++)
        {
            for(j=1;j<=tmp1;j++)
                printf("*");
            for(j=tmp1+1;j<=tmp2-1;j++)
                printf("D");
            for(j=tmp2;j<=n;j++)
                printf("*");
            tmp1++;tmp2--;
            printf("
");
        }
    }
    return 0;
}

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[100010];

int main()
{
    int n,m,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1; i<=n; i++)
        {
            scanf("%d",&str[i]);
        }
        int sum=0,tmp;
        for(i=2; i<=n; i++)
        {
            if(str[i]<str[i-1])
            {
                sum++;
                tmp=i;
            }
                
        }
        if(sum == 0)printf("0
");
        else if(sum > 1)printf("-1
");
        else if(sum == 1 && str[n] <= str[1])
            printf("%d
",n-tmp+1);
        else printf("-1
");
    }
    return 0;
}

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……