Codeforces Round #259 (Div. 2) A. Little Pony and Crystal Mine
2014-08-02 03:14
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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
input
output
input
output
题意:像sample那样输出一个图形。。
思路:这不是语言练习题么。。当时没时间了,没有仔细思考,胡乱写的,发现错位稍微改一下就对了。。
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***
题意:像sample那样输出一个图形。。
思路:这不是语言练习题么。。当时没时间了,没有仔细思考,胡乱写的,发现错位稍微改一下就对了。。
#include <iostream> #include <stdio.h> #include <cmath> #include <algorithm> #include <iomanip> #include <cstdlib> #include <string> #include <memory.h> #include <vector> #include <queue> #include <stack> #include <map> #include <set> #define ll long long #define INF 1000000 using namespace std; int main(){ int n; while(cin>>n){ for(int i=1;i<=n;i++){ int tmp=i; if(i>n/2+1)tmp=n+1-i; for(int j=1;j<=n;j++){ if(j>=n/2+1-(tmp-1)&&j<=n/2+1+(tmp-1)){ printf("D"); }else printf("*"); } cout<<endl; } } return 0; }
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