Codeforces Round #259 (Div. 2) A B C Three consecutive rounds_html/css_WEB-ITnose

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

output

*D*DDD*D*

input

output

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

input

output

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

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;char mapp[110][110];int main(){    int i,j;    int cas;    while(~scanf("%d",&cas))    {        memset(mapp,0,sizeof(mapp));        int ans=cas/2;        for(i=0;i<cas for if>cas-ans-1)                mapp[i][j]='*';            else mapp[i][j]='D';        }        ans--;        cout=0;i--)            cout  <p></p>  <p><br> </p>  <p></p>  <p class="sycode">   </p>
<p class="sycode">    </p>
<p class="sycode">     </p>
<p class="sycode">      </p>
<p class="sycode">       </p>
<p class="sycode">        </p>
<p class="sycode">         B. Little Pony and Sort by Shift        </p>        <p class="sycode">         </p>
<p class="sycode">          time limit per test         </p> 1 second                <p class="sycode">         </p>
<p class="sycode">          memory limit per test         </p> 256 megabytes                <p class="sycode">         </p>
<p class="sycode">          input         </p> standard input                <p class="sycode">         </p>
<p class="sycode">          output         </p> standard output                      <p class="sycode">        </p>
<p> 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:</p> a1,?a2,?...,?an?→?an,?a1,?a2,?...,?an?-?1.        <p> Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?</p>              <p class="sycode">        </p>
<p class="sycode">         Input        </p>        <p> The first line contains an integer n (2?≤?n?≤?105). The second line contains n integer numbers a1,?a2,?...,?an (1?≤?ai?≤?105).</p>              <p class="sycode">        </p>
<p class="sycode">         Output        </p>        <p> If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.</p>              <p class="sycode">        </p>
<p class="sycode">         Sample test(s)        </p>        <p class="sycode">         </p>
<p class="sycode">          </p>
<p class="sycode">           input          </p>          <pre style="代码" class="precsshei">22 1

output

input

31 3 2

output

-1

input

21 2

output

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>int a[100010];int main(){    int i,j;    int cas;    int flag=0;    int ans=0;    while(~scanf("%d",&cas))    {        for(i=0;i<cas scanf for if ans break printf else return>  <br>  <p class="sycode">   </p>
<p class="sycode">    C. Little Pony and Expected Maximum   </p>   <p class="sycode">    </p>
<p class="sycode">     time limit per test    </p> 1 second      <p class="sycode">    </p>
<p class="sycode">     memory limit per test    </p> 256 megabytes      <p class="sycode">    </p>
<p class="sycode">     input    </p> standard input      <p class="sycode">    </p>
<p class="sycode">     output    </p> standard output       <p class="sycode">   </p>
<p> Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.</p>   <p> The dice has m faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the m-th face contains mdots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice n times.</p>    <p class="sycode">   </p>
<p class="sycode">    Input   </p>   <p> A single line contains two integers m and n (1?≤?m,?n?≤?105).</p>    <p class="sycode">   </p>
<p class="sycode">    Output   </p>   <p> Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10??-?4.</p>    <p class="sycode">   </p>
<p class="sycode">    Sample test(s)   </p>   <p class="sycode">    </p>
<p class="sycode">     </p>
<p class="sycode">      input     </p>     <pre style="代码" class="precsshei">6 1

output

3.500000000000

input

6 3

output

4.958333333333

input

2 2

output

1.750000000000

Note

Consider the third test example. If you've made two tosses:

1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2.
2. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1.
3. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2.
4. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.

The probability of each outcome is 0.25, that is expectation equals to:

You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<algorithm>using namespace std;int a[100010];typedef double LL;double qpow(double a,int b){    LL ans=1;    while(b)    {        if(b&1) ans*=a;        b>>=1;        a*=a;    }    return ans;}int main(){    int n;    int m;    scanf("%d%d",&n,&m);    double ans=0;    for(int i=1;i  <br>  <br>  <br> </algorithm></string></cstring></cmath></cstdio></iostream>