A. Little Pony and Crystal Mine

time limit per test

memory limit per test

input

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***

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水題，推出公式，直接打印即可。

代碼：

#include #include using namespace std;int main(){ int n; scanf("%d", &n); for(int i = 1; i <= n; ++i) { if(i <= n/2) { int t = (n+1-2*i)/2; for(int j = 0;j< t; ++j) printf("*"); for(int j = 0; j < 2*i-1; ++j) printf("D"); for(int j=0; j

B. Little Pony and Sort by Shift

time limit per test

memory limit per test

input

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:

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

22 1

output

input

31 3 2

output

-1

input

21 2

output

給一個序列，每次可以把序列的最后一個數移到最前面，如果可以使序列遞增（嚴格來說是非遞減），輸入最少移動次數，否則，輸出-1;

解題思路：

先將序列掃一遍，遇到非遞增的位置（記為t）跳出，接著從從t+1開始掃一遍，如果，后面的序列遞增，則輸出其長度，即為答案，否則，輸出-1。因為如果要按照題中所述方式移動，使序列變為遞增，序列應當是本來就是遞增的，或者是可以分為兩個連續的遞增子序列的。

傳送門：點擊打開鏈接

代碼：

#include #include #include using namespace std;const int MAXN = 1e5 + 10;int n, a[MAXN];int main(){ scanf("%d", &n); for(int i = 0; i < n; ++i) scanf("%d", &a[i]); bool flag = true; int ans = 0, t = 0, i; for(i = 0; i < n-1; ++i) { t = i; if(a[i] > a[i+1]) break; } if(i != n-1) { for(int i = t+1; i < n; ++i) { ans++; if(a[i] > a[(i+1)%n]) { flag = false; break; } } } if(flag) printf("%d\n", ans); else printf("-1\n"); return 0;}

C. Little Pony and Expected Maximum

time limit per test

memory limit per test

input

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

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解題思路：

求數學期望，公式P = m - (1/m)^n - (2/m)^n - ... -((m-1)/m)^n

代碼：

#include #include #include int main(){ int m, n; scanf("%d%d", &m, &n); double ans = m; for(int i = 1; i < m; ++i) ans -= pow(i*1.0/m, n); printf("%.12f\n", ans); return 0;}

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