2018年第九届C/C++ A组蓝桥杯省赛真题——第三题:字母阵列

本文介绍了一个蓝桥杯C/C++比赛题目,挑战者需要在一个100x100的字母矩阵中找出所有隐藏的LANQIAO序列,并提供了具体的实现代码。

2018年第九届C/C++ A组蓝桥杯省赛真题——第三题:字母阵列

如题:
题目描述
仔细寻找,会发现:在下面的8x8的方阵中,隐藏着字母序列:“LANQIAO”。
SLANQIAO
ZOEXCCGB
MOAYWKHI
BCCIPLJQ
SLANQIAO
RSFWFNYA
XIFZVWAL
COAIQNAL

我们约定: 序列可以水平,垂直,或者是斜向;
并且走向不限(实际上就是有一共8种方向)。
上图中一共有4个满足要求的串。

下面有一个更大的(100x100)的字母方阵。
你能算出其中隐藏了多少个“LANQIAO”吗?

FOAIQNALWIKEGNICJWAOSXDHTHZPOLGYELORAUHOHCZIERPTOOJUITQJCFNIYYQHSBEABBQZPNGYQTCLSKZFCYWDGOAIADKLSNGJ
GSOZTQKCCSDWGUWAUOZKNILGVNLMCLXQVBJENIHIVLRPVVXXFTHQUXUAVZZOFFJHYLMGTLANQIAOQQILCDCJERJASNCTLYGRMHGF
TSDFYTLVIBHKLJVVJUDMKGJZGNNSTPVLCKTOFMUEUFSVQIAUVHNVFTGBDDARPKYNNCUOYUAZXQJNOEEYKLFRMOEMHUKJTPETHLES
FKVINSLWEVGAGMKVFVIUBMYOIAFHLVNKNTYKTZWVXQWRWIGPENFXYDTKRVPKRTHMGHVYOCLDCKMEKRLGEKBYUCLOLYPAKPFSOREH
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XWMYRTAFGFOCNHLBGNGOXPJSTWLZUNNAGIRETGXFWAQSSJPFTQAXMTQWMZWYVEPQERKSWTSCHSQOOBGXAQTBCHOEGBDVKGWJIFTG
ZWWJEIISPLMXIMGHOOGDRZFTGNDDWDWMNUFWJYJGULPHNUFSAQNNIUVAAFZIAZKFXXNWCEABGJAUMGYEIEFVQXVHHHEDYUITRCQB
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我就把这些东西放在了txt文件里,读取他们用ifsteam;

ifsteam链接在:这里

其他我就直接取得中间的Q,一条直线来看的

#include <bits/stdc++.h>
using namespace std;
int main()
{
	char a[100][100];
    ifstream fin("1.txt");//我把那些字符放在了1.txt里,读取它 
    for(int i = 0; i < 100; i++)
	{
        for(int j = 0; j < 100; j++)
		{
            fin>>a[i][j];
        }
    }
    fin.close();//关闭读取文件 
	int sum,i,j;
	for(i=0;i<105;i++)//把Q放在前面,分成八个方向分别判断是否成为LANQIAO。 
	{
		for(j=0;j<105;j++)
		{
	 		if(a[i][j]=='Q'&&j>=3&&j<97)
	 		{
	  			if(a[i][j-3]=='L'&&a[i][j-2]=='A'&&a[i][j-1]=='N'&&a[i][j+1]=='I'&&a[i][j+2]=='A'&&a[i][j+3]=='O')
	  			sum++;
	  
	  			if(a[i][j-3]=='O'&&a[i][j-2]=='A'&&a[i][j-1]=='I'&&a[i][j+1]=='N'&&a[i][j+2]=='A'&&a[i][j+3]=='L')
	  			sum++;
	  		}
	 
	 		if(a[i][j]=='Q'&&i>=3&&i<97)
			{
		  		if(a[i-3][j]=='L'&&a[i-2][j]=='A'&&a[i-1][j]=='N'&&a[i+1][j]=='I'&&a[i+2][j]=='A'&&a[i+3][j]=='O')
		  		sum++;
		  		if(a[i-3][j]=='O'&&a[i-2][j]=='A'&&a[i-1][j]=='I'&&a[i+1][j]=='N'&&a[i+2][j]=='A'&&a[i+3][j]=='L')
		  		sum++;
	  		}
	 
		 	if(a[i][j]=='Q'&&i>=3&&i<97&&j>=3&&j<97)
			{
		  		if(a[i-3][j-3]=='L'&&a[i-2][j-2]=='A'&&a[i-1][j-1]=='N'&&a[i+1][j+1]=='I'&&a[i+2][j+2]=='A'&&a[i+3][j+3]=='O')
		  		sum++;
		  		if(a[i-3][j-3]=='O'&&a[i-2][j-2]=='A'&&a[i-1][j-1]=='I'&&a[i+1][j+1]=='N'&&a[i+2][j+2]=='A'&&a[i+3][j+3]=='L')
		  		sum++;
		  		if(a[i+3][j-3]=='L'&&a[i+2][j-2]=='A'&&a[i+1][j-1]=='N'&&a[i-1][j+1]=='I'&&a[i-2][j+2]=='A'&&a[i-3][j+3]=='O')
		  		sum++;
		  		if(a[i+3][j-3]=='O'&&a[i+2][j-2]=='A'&&a[i+1][j-1]=='I'&&a[i-1][j+1]=='N'&&a[i-2][j+2]=='A'&&a[i-3][j+3]=='L')
		  		sum++;
		  	}
		}
	}
	printf("%d",sum);
}

在这里插入图片描述
答案:41

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