C. Chessboard( Educational Codeforces Round 41 (Rated for Div. 2))
//暴力
#include <iostream>
#include <algorithm>
#include <string> using namespace std;
const int N = ;
string s1[N], s2[N], s3[N], s4[N];
int a[N][N], b[N][N]; int main()
{
int n;
cin >> n;
for (int i = ; i<n; i++)
cin >> s1[i];
cin.get();
for (int i = ; i<n; i++)
cin >> s2[i];
cin.get();
for (int i = ; i<n; i++)
cin >> s3[i];
cin.get();
for (int i = ; i<n; i++)
cin >> s4[i]; for (int i = ; i<n; i++)
for (int j = ; j < n; j++){
//只有两种方式
a[i][j] = (i + j) % == ? : ;
b[i][j] = (i + j) % ? : ;
}
for (int i = ; i<n; i++)
for (int j = ; j<n; j++)
a[i][j] += , b[i][j] += ; //比较
int ans = 1e9;
int f = ;
for (int i = ; i<n; i++)
for (int j = ; j<n; j++)
f += (s1[i][j] != a[i][j]) + (s2[i][j] != a[i][j]) + (s3[i][j] != b[i][j]) + (s4[i][j] != b[i][j]);
ans = min(ans, f);
f = ;
for (int i = ; i<n; i++)
for (int j = ; j<n; j++)
f += (s1[i][j] != a[i][j]) + (s2[i][j] != b[i][j]) + (s3[i][j] != a[i][j]) + (s4[i][j] != b[i][j]);
ans = min(ans, f);
f = ;
for (int i = ; i<n; i++)
for (int j = ; j<n; j++)
f += (s1[i][j] != a[i][j]) + (s2[i][j] != b[i][j]) + (s3[i][j] != b[i][j]) + (s4[i][j] != a[i][j]);
ans = min(ans, f);
f = ;
for (int i = ; i<n; i++)
for (int j = ; j<n; j++)
f += (s1[i][j] != b[i][j]) + (s2[i][j] != a[i][j]) + (s3[i][j] != a[i][j]) + (s4[i][j] != b[i][j]);
ans = min(ans, f);
f = ;
for (int i = ; i<n; i++)
for (int j = ; j<n; j++)
f += (s1[i][j] != b[i][j]) + (s2[i][j] != a[i][j]) + (s3[i][j] != b[i][j]) + (s4[i][j] != a[i][j]);
ans = min(ans, f);
f = ;
for (int i = ; i<n; i++)
for (int j = ; j<n; j++)
f += (s1[i][j] != b[i][j]) + (s2[i][j] != b[i][j]) + (s3[i][j] != a[i][j]) + (s4[i][j] != a[i][j]);
ans = min(ans, f);
cout<< ans <<endl;
system("pause");
return ;
}
C. Chessboard( Educational Codeforces Round 41 (Rated for Div. 2))的更多相关文章
- D. Pair Of Lines( Educational Codeforces Round 41 (Rated for Div. 2))
#include <vector> #include <iostream> #include <algorithm> using namespace std; ty ...
- B. Lecture Sleep( Educational Codeforces Round 41 (Rated for Div. 2))
前缀后缀和搞一搞,然后枚举一下区间,找出最大值 #include <iostream> #include <algorithm> using namespace std; ; ...
- (模拟)关于进制的瞎搞---You Are Given a Decimal String...(Educational Codeforces Round 70 (Rated for Div. 2))
题目链接:https://codeforc.es/contest/1202/problem/B 题意: 给你一串数,问你插入最少多少数可以使x-y型机器(每次+x或+y的机器,机器每次只取最低位--% ...
- Educational Codeforces Round 41 (Rated for Div. 2)(A~D)
由于之前打过了这场比赛的E题,而后面两道题太难,所以就手速半个多小时A了前4题. 就当练手速吧,不过今天除了C题数组开小了以外都是1A A Tetris 题意的抽象解释可以在Luogu里看一下(话说现 ...
- Multidimensional Queries(二进制枚举+线段树+Educational Codeforces Round 56 (Rated for Div. 2))
题目链接: https://codeforces.com/contest/1093/problem/G 题目: 题意: 在k维空间中有n个点,每次给你两种操作,一种是将某一个点的坐标改为另一个坐标,一 ...
- Educational Codeforces Round 41 (Rated for Div. 2) ABCDEF
最近打的比较少...就只有这么点题解了. A. Tetris time limit per test 1 second memory limit per test 256 megabytes inpu ...
- Educational Codeforces Round 41 (Rated for Div. 2)
这场没打又亏疯了!!! A - Tetris : 类似俄罗斯方块,模拟一下就好啦. #include<bits/stdc++.h> #define fi first #define se ...
- Educational Codeforces Round 41 (Rated for Div. 2) D. Pair Of Lines (几何,随机)
D. Pair Of Lines time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...
- Educational Codeforces Round 78 (Rated for Div. 2)E(构造,DFS)
DFS,把和当前结点相连的点全都括在当前结点左右区间里,它们的左端点依次++,然后对这些结点进行DFS,优先对左端点更大的进行DFS,这样它右端点会先括起来,和它同层的结点(后DFS的那些)的区间会把 ...
随机推荐
- BZOJ3627: [JLOI2014]路径规划
BZOJ3627: [JLOI2014]路径规划 Description 相信大家都用过地图上的路径规划功能,只要输入起点终点就能找出一条最优路线.现在告诉你一张地图的信息,请你找出最优路径(即最短路 ...
- 20170316 ABAP注意点
1.debug 时在MODIFY db from table 后数据便提交了: 一般情况下,更新数据库需要commit,但debug会自动commit,程序结束也会自动commit. 2.使用at n ...
- [haoi2015]T1
题意:给定你一颗树,要求你在这棵树中确定K个黑点和N-K个白点,使黑点间与白点间两两距离之和最大,输出最大值.n<=2000 对于这道题,我想了好几个思路,包括点分治,贪心,动规,网络流等等,实 ...
- ansible操作模块相关
1. 查看模块可用参数命令 ansible-doc -s module_name
- zkui部署
1.拉取代码 #git clone https://github.com/DeemOpen/zkui.git 2.构建并安装程序 #cd zkui/ #yum install -y maven #mv ...
- CentOS系统文件和目录管理相关的一些重要命令
我们都知道,在Linux系统中,基本上任何我们需要做的事都可以通过输入命令来完成,所以在Linux系统中命令非常的多,我们不可能也没必要记住所有的这些命令,但是对于一些常用的命令我们还是必须要对其了如 ...
- (QA-LSTM)自然语言处理:智能问答 IBM 保险QA QA-LSTM 实现笔记.md
train集: 包含若干条与保险相关的问题,每一组问题对为一行,示意如下: 可分为四项,第三项为问题,第四项为答案: 1.build_vocab 统计训练集中出现的词,返回结果如下(一个包含3085个 ...
- jni中c代码调用java代码
原理是使用反射的机制 java中反射的例子: Class<?> forName = Class.forName("com.example.ndkcallback.DataProv ...
- 机器学习之K-means算法
前言 以下内容是个人学习之后的感悟,转载请注明出处~ 简介 在之前发表的线性回归.逻辑回归.神经网络.SVM支持向量机等算法都是监督学习算法,需要样本进行训练,且 样本的类别是知 ...
- ZipHelper
using ICSharpCode.SharpZipLib.Zip; using System.Collections.Generic; using System.IO; namespace WLYD ...