第十一届蓝桥杯回文日期 Python题解

Author： artrajz
发布时间：March 10, 2022
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Categories：题解

## 题目

![](https://s1.ax1x.com/2022/03/10/b4e9mQ.png)

![](https://s1.ax1x.com/2022/03/10/b4eFkn.png)

## 思路

因为是回文日期，所以没必要遍历每一天，直接用年份造一个回文日期再判断它是不是一个正确的日期就行了。

吐槽一下这个评测用例的范围，明明说的是10000101<=N<=89991231,于是我设置的最大年份是9000年，结果一直卡在90分，找了半天也没发现问题。最后在机缘巧合之下把范围调成最大9999年就100了，题目范围害人不浅啊。。

## 题解

```Python
n = str(input())
date1 = 0
date2 = 0

days = [31,28,31,30,31,30,31,31,30,31,30,31]
def isLeap(year):#判断是否是闰年 
    if (year % 4 == 0 and year % 100 != 0) or year % 400 == 0:
        return True
    else:
        return False
    
def isDate(date):#判断是否是正确的日期
    flag = False
    if int(date[4:6]) > 0 and int(date[4:6]) <= 12:
        if int(date[4:6]) == 2 and isLeap(int(date[0:4])):
            if int(date[6:8]) > 0 and int(date[6:8]) <= 29:
                flag = True
        elif int(date[6:8]) > 0 and int(date[6:8]) <= days[int(date[4:6]) - 1]:
            flag = True
    return flag
    
for i in range(int(n[0:4]),9999):
    date = str(i) + str(i)[::-1]
    if isDate(date) and int(date) > int(n):
        date1 = date
        break
    
for i in range(int(n[0:4]),9999):
    date = str(i) + str(i)[::-1]
    if isDate(date) and str(i)[0:2] == str(i)[2:4] and int(date) > int(n):
        date2 = date
        break
    
print(date1)
print(date2)
```

![评测结果](https://s1.ax1x.com/2022/03/10/b4eLB4.jpg)

Last modification：March 11, 2022