dic1={} for i in range(1000,5000,1): n=100000//i str1=str(i) for k in range(2,n,1): if str(i*k)==reversed(str1): dic1[i]=k break print(dic1)请修改以上代码
时间: 2024-03-10 22:46:52 浏览: 13
以下是修改后的代码:
```
dic1 = {}
for i in range(1000, 5000):
n = 100000 // i
str1 = str(i)
for k in range(2, n):
if str(i * k) == str(i * k)[::-1]:
dic1[i] = k
break
print(dic1)
```
修改说明:
1. 修改了 `range` 函数的范围,使其不包含5000。
2. 修改了判断回文数的方式,使用了字符串切片反转的方法,即 `[::-1]`。
3. 将 `range` 函数中的步长从1改为默认值1,这是因为步长的默认值就是1,不需要显式地指定。
相关问题
dic1={} for i in range(1000,5000,1): n=10000//i a=0 for k in range(2,n,1): if i*k==int(str(i)[::-1]): if a==0: dic1[i]=[] dic1[i].append(k) a=a+1 else: dic1[i].append(k) for key,value in dic1.items(): print(str(key)+':'+str(value))请修改
修改后的代码如下:
```
dic1 = {}
for i in range(1000, 5000, 1):
n = 10000 // i
for k in range(2, n):
if i * k == int(str(i * k)[::-1]):
if i not in dic1:
dic1[i] = []
dic1[i].append(k)
for key, value in dic1.items():
print(str(key) + ':' + str(value))
```
修改说明:
1. 修改了循环的范围,将 `range(2,n,1)` 改为 `range(2,n)`;
2. 在判断条件中,加上了 `i*k` 的回文数判断;
3. 在字典赋值时,判断了当前的 `i` 是否已经在字典中,如果不在就初始化一个空列表。
这段代码报KeyError错误,该如何修改:def oddString(self, words: List[str]) -> str: n = len(words) dic = {} ans = 0 for i in range(0, n): len_word = len(words[i]) diff = [ord(words[i][j]) - ord(words[i][j-1]) for j in range(1, len_word)] if tuple(diff) in dic.keys(): dic[tuple(diff)] = 0 else: dic[tuple(diff)] += 1 if dic[tuple(diff)] == 1: ans = 1 return words[ans]
这段代码会报KeyError错误,是因为在if条件语句中,当dic字典中不存在tuple(diff)这个键时,会执行dic[tuple(diff)] += 1语句,但是此时dic字典中并没有tuple(diff)这个键,所以会报KeyError错误。
为了避免这个错误,可以在if条件语句前先判断一下tuple(diff)是否已经在字典中存在,如果不存在,就先将其添加到字典中,然后再执行后面的代码。修改后的代码如下:
```
def oddString(self, words: List[str]) -> str:
n = len(words)
dic = {}
ans = 0
for i in range(0, n):
len_word = len(words[i])
diff = [ord(words[i][j]) - ord(words[i][j-1]) for j in range(1, len_word)]
if tuple(diff) not in dic:
dic[tuple(diff)] = 0
if dic[tuple(diff)] == 0:
ans = i
dic[tuple(diff)] += 1
return words[ans]
```
这里将if条件语句中的dic[tuple(diff)] == 1改为了dic[tuple(diff)] == 0,同时将ans的初始值改为0,这样就可以正确地找到奇异字符串了。