思路：
每一列都整成一個串列，然后全排列
a = [97,92,0,0,89,82,0,0,0,95,0,0,94,0,0,0,98,93,0,0]
b = [90,85,0,0,83,86,0,97,0,99,0,0,91,83,0,0,83,87,0,99]
c = [0,96,0,0,97,0,0,96,89,0,96,0,0,87,98,0,99,92,0,96]
d = [0,0,0,80,0,0,87,0,0,0,97,93,0,0,97,93,98,96,89,95]
e = [0,0,93,86,0,0,90,0,0,0,0,98,0,0,98,86,81,98,92,81]
ans = 0
for i in range(20):
for j in range(20):
if i == j:
continue
for k in range(20):
if i == k or j == k:
continue
for l in range(20):
if i == l or j == l or k == l:
continue
for m in range(20):
if i == m or j == m or k == m or l == m:
continue
sum1 = a[i]+b[j]+c[k]+d[l]+e[m]
if sum1 > ans:
ans = sum1
print(ans)
思路：
計算器直接算，步驟如下：
2019 / 26 = 77 2019 % 26 = 17=Q
77 / 26 = 2 77 % 26 = 25=Y
2=B
最終答案：BYQ
思路：
斐波那契數列相信大家都不陌生，這里面需要注意的還是超出資料范圍的問題，最后需要保留四位小數，只需要每次求出的和對10000取余即可，
a = 1
b = 1
c = 3
temp = 0
for i in range(4,20190324):
d = (a + b + c) % 10000
a = b
b = c
c = d
print(int(d))
思路：
暴力三層for回圈，需要特別注意的是，最后結果要除以6
def check(n):
while n > 10:
i = n%10
if i == 2 or i == 4:
return 0
n = n//10
return 1
sum = 0
num = 0
for i in range(1,2019):
for j in range(1,2019):
for k in range(1, 2019):
if k == j and i == j and i == k:
sum = i + j + k
if sum > 2019:
break
if check(i) == 1 and check(j) == 1 and check(k) == 1 and sum == 2019:
num += 1
sum = 0
print(num/6)
思路：
跟D題很相似
j = 0
a, b = 0, 0
n = int(input())
for i in range(1,n+1):
j = i
while j != 0 :
if j % 10 == 0 or j % 10 == 2 or j % 10 == 1 or j % 10 == 9:
b = 1
break
j = j // 10
if b == 1:
a = a + i
b = 0
print(a)
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標籤：python
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