2018第九届蓝桥杯JAVA C组 第三题 字母阵列 题解

标题：字母阵列

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
KWPUOLOVMOFBIXYACWRDYBINTMPASPCEOKHXQIGBQQMCEOIVULIEOPFSCSIHENAJCVDPJDOIWIIULFDNOZOFVAMCABVGKAKCOZMG
XWMYRTAFGFOCNHLBGNGOXPJSTWLZUNNAGIRETGXFWAQSSJPFTQAXMTQWMZWYVEPQERKSWTSCHSQOOBGXAQTBCHOEGBDVKGWJIFTG
ZWWJEIISPLMXIMGHOOGDRZFTGNDDWDWMNUFWJYJGULPHNUFSAQNNIUVAAFZIAZKFXXNWCEABGJAUMGYEIEFVQXVHHHEDYUITRCQB
XZHDPZQTOBECJVBZLACVXACZEDYOGVAVQRNWEOWGRAQYUEUESTEDQTYJUTEFOOITSHDDZHONJGBRCWNEQLZUTBNQIADKNFIOMWZR
EBFKCVNLURZSNPOLTISRPDTNUMCDGKTYRGIOVEPTUTSBAWQKWWEUWIWHAANUZUADGZEATZOQICWFUJTWNZDBKLQNELWVTBNDLNFH
PESISEATZNCDFRMXBQUKBFTIGYSFCWVHPMSUSDKPSCOMVLDOHYQVFHAJKRDTAVLIMNZBZSMLMRTLRPSLAHXDBASDMWAAYBPYVJZF
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JCAULIQRYUMAMAOVVWSEUTMECXSDTONRMMROQUISYEURSAYNZUVOPXLIFBDOHPXMABBLEQZGLJXQJOEYYRRRCFTEZQAOIWKRJQDL
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MVBNFXMFNPAYSODPXEAYNRKTEZJWMUACSIUYPIORUFPMXAOZZJPJXPFLNSKNIAMETMOVULZPQIJJIRCSYQXOEVRHCNACSBRHKYNW
KGKBTBHGWKVJYZCOVNSKUREKZEIWVLOHAMUAYKLUGHEUESICBZAHURNTJAECTHRNKSIJQFIPVZANSZYSPJWHPKHCAPEYWNXUYQSD
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BORWTNBISSLTVKBRTLWKRNCEDCNEGCIYJIPDICFAVNOISYAHWBLGMNFKXZYTTWJOBEPNMSJEJMHXVPGOJOLQQQVXFGEULANQIAOD
OQETOJHCZXGTUKIWGMEVVMXCURISUOFQSAWZWDMZWVYHZMPEIMWKJDGERODVVUXYRTYLCRGYQQOIOFZSSZRAIESWBQOAIQNALJNR
HEYWHPLLPCUEOCBAOWGAYEJZQJHLVNMVQNSQQGGUBOIMDPFLOVSQGBLYAMBRYJDVOXOQINLJAVYALAKHPKOYNKGXIISSJNGKHYMS
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PDOFSCKKJJTRYRWGLEZODQTJSIMXIAOLNMLPHBAYLPTTLPYWILSEIIQVSXNHIJEORVCNJHYXRBIZZJTADGMRTSXVRXYGVQQNUEIC
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FBQASGNCCBBGNKJHCDBTGBIIWKMPHDABKEWDEPYEAVKNMPATUZZUOEHGUGAZNECSGUCIIJPMMRAMTVADMTCRJCBWDLWWFNFOWMVZ
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CURAGWXMLYBSIZLAXFONZZMQMRNNSRQKRHQGFGZUTLONAYRKSSOWAMKZBSGOOYQDPTBHGPBNQEDCZHRTOXREOFJEKJVIZXZBCJPN
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DDAOBVCPOVODVJFLSNPJXHWQBHILWZAHQQMTQASNADZLZNXJLJMFCOUWOZJCMVVTYCKTUBABWLCEBNYWAMOLNBQQYBRUJCQCZALE
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DCZGKILMBFXIQQFEKJBIUDEMIFCANVGNYZAYSQFMNNQFEPZFUUVGTBKSMDXITBLANQIAOQUKTPNYPOWSQQYWWMJHSDYVFDJYXBAF
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SYSFUAJFKIPUJVQSZRWQCJYAUMLDDNOKODDXIEQIFLANQIAOZFUNKUBVDBLMJOAUTVCZVLKJRQIORQPGAVCEYVNYUZHXILHERYEC
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WSDAUGUTVWCVHEMOIRJJGTANUWTSAIXXEVZTBDHPGSRHHVWCDZVZYRJTLONIJVXEATHQXOUKBIGZONFRSZIOGWNTYAJYLQCGEOWY

请提交一个整数，不要填写任何多余的内容。

解题思路：我是用C语言解的纯暴力，JAVA解法也一样。先将这堆字母存进数组，然后遍历，看见‘L’就找他周围的‘A’，‘N’，‘Q’，‘I’，‘A’，‘O’。找到O就结果s++，最后输出。为了验证，我加了个标记数组f[100][100]。找到O就将LANQIAO全部赋值为1最后输出很可观

答案是：41。

#include
char a[100][100];
int f[100][100]={0};
int main(){
	for(int i=0;i<100;i++)
		scanf("%s",a[i]);
	int s=0;
	for(int i=0;i<100;i++){
		for(int j=0;j<100;j++){
			if(a[i][j]=='L'&&i+6<100&&j+6<100){
				if(a[i+1][j+1]=='A'){
					if(a[i+2][j+2]=='N'){
						if(a[i+3][j+3]=='Q'){
							if(a[i+4][j+4]=='I'){
								if(a[i+5][j+5]=='A'){
									if(a[i+6][j+6]=='O'){
										f[i][j]=1;
										f[i+1][j+1]=1;
										f[i+2][j+2]=1;
										f[i+3][j+3]=1;
										f[i+4][j+4]=1;
										f[i+5][j+5]=1;
										f[i+6][j+6]=1;
										s++;
									}
								}
							}
						}
					}
				}
			}
			
			if(a[i][j]=='L'&&i+6<100&&j-6>=0){
				if(a[i+1][j-1]=='A'){
					if(a[i+2][j-2]=='N'){
						if(a[i+3][j-3]=='Q'){
							if(a[i+4][j-4]=='I'){
								if(a[i+5][j-5]=='A'){
									if(a[i+6][j-6]=='O'){
										f[i][j]=1;
										f[i+1][j-1]=1;
										f[i+2][j-2]=1;
										f[i+3][j-3]=1;
										f[i+4][j-4]=1;
										f[i+5][j-5]=1;
										f[i+6][j-6]=1;
										s++;
									}
								}
							}
						}
					}
				}
			}
			
			
			if(a[i][j]=='L'&&i-6>=0&&j+6<100){
				if(a[i-1][j+1]=='A'){
					if(a[i-2][j+2]=='N'){
						if(a[i-3][j+3]=='Q'){
							if(a[i-4][j+4]=='I'){
								if(a[i-5][j+5]=='A'){
									if(a[i-6][j+6]=='O'){
										f[i][j]=1;
										f[i-1][j+1]=1;
										f[i-2][j+2]=1;
										f[i-3][j+3]=1;
										f[i-4][j+4]=1;
										f[i-5][j+5]=1;
										f[i-6][j+6]=1;
										s++;
									}
								}
							}
						}
					}
				}
			}
			
			if(a[i][j]=='L'&&i-6>=0&&j-6>=0){
				if(a[i-1][j-1]=='A'){
					if(a[i-2][j-2]=='N'){
						if(a[i-3][j-3]=='Q'){
							if(a[i-4][j-4]=='I'){
								if(a[i-5][j-5]=='A'){
									if(a[i-6][j-6]=='O'){
										f[i][j]=1;
										f[i-1][j-1]=1;
										f[i-2][j-2]=1;
										f[i-3][j-3]=1;
										f[i-4][j-4]=1;
										f[i-5][j-5]=1;
										f[i-6][j-6]=1;
										s++;
									}
								}
							}
						}
					}
				}
			}
			
			if(a[i][j]=='L'&&j-6>=0){
				if(a[i][j-1]=='A'){
					if(a[i][j-2]=='N'){
						if(a[i][j-3]=='Q'){
							if(a[i][j-4]=='I'){
								if(a[i][j-5]=='A'){
									if(a[i][j-6]=='O'){
										f[i][j]=1;
										f[i][j-1]=1;
										f[i][j-2]=1;
										f[i][j-3]=1;
										f[i][j-4]=1;
										f[i][j-5]=1;
										f[i][j-6]=1;
										s++;
									}
								}
							}
						}
					}
				}
			}
			
			if(a[i][j]=='L'&&j+6<100){
				if(a[i][j+1]=='A'){
					if(a[i][j+2]=='N'){
						if(a[i][j+3]=='Q'){
							if(a[i][j+4]=='I'){
								if(a[i][j+5]=='A'){
									if(a[i][j+6]=='O'){
										f[i][j]=1;
										f[i][j+1]=1;
										f[i][j+2]=1;
										f[i][j+3]=1;
										f[i][j+4]=1;
										f[i][j+5]=1;
										f[i][j+6]=1;
										s++;
									}
								}
							}
						}
					}
				}
			}
			
			if(a[i][j]=='L'&&i-6>=0){
				if(a[i-1][j]=='A'){
					if(a[i-2][j]=='N'){
						if(a[i-3][j]=='Q'){
							if(a[i-4][j]=='I'){
								if(a[i-5][j]=='A'){
									if(a[i-6][j]=='O'){
										f[i][j]=1;
										f[i-1][j]=1;
										f[i-2][j]=1;
										f[i-3][j]=1;
										f[i-4][j]=1;
										f[i-5][j]=1;
										f[i-6][j]=1;
										s++;
									}
								}
							}
						}
					}
				}
			}
			
			if(a[i][j]=='L'&&i+6<100){
				if(a[i+1][j]=='A'){
					if(a[i+2][j]=='N'){
						if(a[i+3][j]=='Q'){
							if(a[i+4][j]=='I'){
								if(a[i+5][j]=='A'){
									if(a[i+6][j]=='O'){
										f[i][j]=1;
										f[i+1][j]=1;
										f[i+2][j]=1;
										f[i+3][j]=1;
										f[i+4][j]=1;
										f[i+5][j]=1;
										f[i+6][j]=1;
										s++;
									}
								}
							}
						}
					}
				}
			}
			
		}
	}
	for(int i=0;i<100;i++){
		for(int j=0;j<100;j++){
			if(f[i][j]==1)
				printf("%c",a[i][j]);
			else
				printf(" ");
		}
		printf("\n");
	}
				 printf("\n\n%d",s);

	return 0;
}