[TC] SRM 633 div2 250

最新推荐文章于 2020-04-13 18:58:22 发布

Modiz

最新推荐文章于 2020-04-13 18:58:22 发布

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分类专栏：基础 CF、TC 文章标签：模拟

本文链接：https://blog.csdn.net/modiz/article/details/41877073

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本文指导您如何绘制一个由内向外嵌套的正方形组成的靶子图案，具体步骤涉及不同大小的正方形使用空格和井号字符交错排列。

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Problem Statement

Here at [topcoder], we call a contestant a "target" if their rating is 3000 or more. In the arena, the targets have a red icon with a small target on it. Do you want to become a target as well? Sure you do. But before you get there, let's start with something easier: drawing a target.

The target you need to draw consists of nested squares. The innermost square is just a single '#' character. The larger squares use alternatingly the character ' ' (space) and the character '#'. Here is an example in which the side of the largest square is n = 5:

#####
#   #
# # #
#   #
#####

And here is an example for n = 9:

#########
#       #
# ##### #
# #   # #
# # # # #
# #   # #
# ##### #
#       #
#########

You will be given an int n. Your method must return a vector <string> which contains a drawing of the target with side n. More precisely, each element of the returned vector <string> must be one row of the drawing, in order. Therefore, the returned vector <string> will consist of n elements, each with n characters. (See the examples below for clarification.)

The value of n will be such that a target like the ones above can be drawn: 5, 9, 13, and so on. Formally, n will be of the form 4k+1, where k is a positive integer.

Definition

Class:	Target
Method:	draw
Parameters:	int
Returns:	vector <string>
Method signature:	vector <string> draw(int n)
(be sure your method is public)

Limits

Time limit (s):	2.000
Memory limit (MB):	256

Constraints

n will be between 5 and 49, inclusive.

n mod 4 will be 1.

Examples

Returns: {"#####", "#   #", "# # #", "#   #", "#####" }

Returns: 
{"#########",
 "#       #",
 "# ##### #",
 "# #   # #",
 "# # # # #",
 "# #   # #",
 "# ##### #",
 "#       #",
 "#########" }

Returns: 
{"#############",
 "#           #",
 "# ######### #",
 "# #       # #",
 "# # ##### # #",
 "# # #   # # #",
 "# # # # # # #",
 "# # #   # # #",
 "# # ##### # #",
 "# #       # #",
 "# ######### #",
 "#           #",
 "#############" }

Returns: 
{"#################",
 "#               #",
 "# ############# #",
 "# #           # #",
 "# # ######### # #",
 "# # #       # # #",
 "# # # ##### # # #",
 "# # # #   # # # #",
 "# # # # # # # # #",
 "# # # #   # # # #",
 "# # # ##### # # #",
 "# # #       # # #",
 "# # ######### # #",
 "# #           # #",
 "# ############# #",
 "#               #",
 "#################" }

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

class Target
{
        public:
        vector <string> draw(int n)
        {
        	int i,j;
        	VS ans;
            for (i=0;i<n;i++){
            	if (!(i&1)){
            		string s="";
            		int k=min(i/2,(n-1-i)/2);
            		for (j=1;j<=n;j++){
            			if (!(j&1) && ((j<=n/2 && j/2<=k) || (j>n/2) && (n-j+1)/2<=k)) s+=' ';
						else s+='#';
            		}
            		ans.push_back(s);
            	}
            	else{
            		string s="";
            		int k=min((i+1)/2,(n-i)/2);
            		for (j=1;j<=n;j++){
            			if ((j&1) && ((j<=n/2 && (j+1)/2<=k) || (j>n/2 && (n-j+2)/2<=k))) s+='#';
            			else s+=' ';
            		}
            		ans.push_back(s);
            	}
            }
            //int k=ans.size();
            //for (i=0;i<k;i++)
            //	cout<<ans[i]<<endl;
            return ans;
        }
    }