2018/Province_Java_C/3/字母阵列
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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KGKBTBHGWKVJYZCOVNSKUREKZEIWVLOHAMUAYKLUGHEUESICBZAHURNTJAECTHRNKSIJQFIPVZANSZYSPJWHPKHCAPEYWNXUYQSD
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NMBKMUENVGXXVMQCLXPJDQIQCFWYADIFDSGINGZDJYHPUPXVRMWDIPJRWPNRYOFGYYPEAVKDEMLYRRRMNCRQXPTDSQIVKKGJWDEF
SBAEKIFZCKDOMIQKBDWVQGBYWPDIBOLQUGAQRXLJDAZMXVZXYSNWEWTNZKYREMBEUHOTFOCKEJSXCMUBCKXNGQXTQJRCRCLWJTOI
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请提交一个整数,不要填写任何多余的内容。
Ideas
就是简单的数据处理+遍历+check。
Code
Python
def check(x: int, y: int) -> bool:res = 0for dx, dy in [(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1)]:tmp = list()for i in range(7):nx, ny = x + i * dx, y + i * dyif -1 < nx < n and -1 < ny < n:tmp.append(maps[nx][ny])else:continueif ''.join(tmp) == "LANQIAO":res += 1return resif __name__ == '__main__':n = 100maps = list()with open("input.txt", 'r') as fp:for line in fp.readlines():maps.append([])for ch in line.strip():maps[-1].append(ch)# print(maps)ans = 0for i in range(n):for j in range(n):ans += check(i, j)print(ans)
Answer:41
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