9.19网易笔试（红蓝格子第一题），第四题只能通过70%。
第一题：
思路：连续颜色算成一块，寻找红色的块数、蓝色的块数，结果就是红蓝中比较小的块数+1。
import sys N = int(sys.stdin.readline().strip()) value = list(sys.stdin.readline().strip()) blueCount = 0 tag = 0 for i in range(N): if value[i] == 'B' and tag == 0: tag = 1 if tag == 1 and value[i] == 'R': tag = 0 blueCount += 1 redCount = 0 tag = 0 for i in range(N): if value[i] == 'R' and tag == 0: tag = 1 if tag == 1 and value[i] == 'B': tag = 0 redCount += 1 if value[-1] == 'R': redCount += 1 else: blueCount += 1 print(min(blueCount, redCount) + 1)第二题：
不能整除返回-1，能整除就分解质因数，结果就是质因数的数目。
import sys def factorize(n): res = [] i = 2 while(i**2 <= n): while n % i == 0: res.append(i) n = n // i i += 1 if n != 1: res.append(n) return len(res) line = sys.stdin.readline().strip() values = list(map(int, line.split())) a = values[0] b = values[1] if a >= b&nbs***bsp;b % a != 0: print(-1) else: print(factorize(b // a))
第三题：时间复杂度很宽松，暴力匹配就可以。
import sys p = sys.stdin.readline().strip() p = list(p) def calculateCost(pattern): p = ['A','c','M','e','r'] summation = 0 for i in range(5): cost = 10 if abs(ord(p[i]) - ord(pattern[i])) == 32: cost -= 5 elif ord(p[i]) == ord(pattern[i]): cost -= 10 elif p[i].isupper() == pattern[i].isupper(): cost -= 5 summation += cost return summation res = 50 for i in range(0,len(p) - 4): pattern = p[i:i+5] summation = calculateCost(pattern) res = min(res, summation) print(res)
第四题：我的思路是求等比数列的通项公式，最后结果是n(q-1)+1 = qk
q从2到n遍历求k即可，只能通过70%，求更好的思路。
import sys import math m = int(sys.stdin.readline().strip()) def calculate(x): for q in range(2, x): res = x*(q-1) + 1 if abs(math.log(res)/math.log(q) - round(math.log(res)/math.log(q))) < 0.00001: print(str(round(math.log(res)/math.log(q))) + ' ' + str(q)) return for i in range(m): value = int(sys.stdin.readline().strip()) calculate(value)
