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***
传送门：点击打开链接
解题思路：水题，推出公式，直接打印即可。
代码：
#include <cstdio>#include <cstring>using namespace std;int main(){ int n; scanf("%d", &n); for(int i = 1; i <br> <p></p> <p class="sycode"> </p>
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<h2>B. Little Pony and Sort by Shift</h2> <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
给一个序列，每次可以把序列的最后一个数移到最前面，如果可以使序列递增（严格来说是非递减），输入最少移动次数，否则，输出-1;
解题思路：
先将序列扫一遍，遇到非递增的位置（记为t）跳出，接着从从t+1开始扫一遍，如果，后面的序列递增，则输出其长度，即为答案，否则，输出-1。因为如果要按照题中所述方式移动，使序列变为递增，序列应当是本来就是递增的，或者是可以分为两个连续的递增子序列的。
传送门：点击打开链接
代码：
#include <cstdio>#include <cstring>#include <queue>using namespace std;const int MAXN = 1e5 + 10;int n, a[MAXN];int main(){ scanf("%d", &n); for(int i = 0; i a[i+1]) break; } if(i != n-1) { for(int i = t+1; i a[(i+1)%n]) { flag = false; break; } } } if(flag) printf("%d\n", ans); else printf("-1\n"); return 0;}</queue></cstring></cstdio>
C. Little Pony and Expected Maximum
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
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.
Input
A single line contains two integers m and n (1?≤?m,?n?≤?105).
Output
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.
Sample test(s)
input
6 1
output
3.500000000000
input
6 3
output
4.958333333333
input
2 2
output
1.750000000000
传送门： 点击打开链接
解题思路：
求数学期望，公式P = m - (1/m)^n - (2/m)^n - ... -((m-1)/m)^n
代码：
#include <cstdio>#include <cstring>#include <cmath>int main(){ int m, n; scanf("%d%d", &m, &n); double ans = m; for(int i = 1; i <br> <br> <br> <p></p> </cmath></cstring></cstdio>
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