Codeforces AIM Tech Round 4 (Div. 2) A. Diversity
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题目网址: Codeforces AIM Tech Round 4 (Div. 2) A. Diversity
分析
- 给你一个字符串s, 求你最少次数变换里面的字符使得里面至少有k个不同的字符
思路
- (为了体验python写acm)
- 判断k是否大于字符串长度,是则 ‘impossible’
- 否则, 计算k-减去字符串中已有的不同字符个数
代码
ipu = input()k = int(input())ipuLen = len(ipu)diffLen = len(set(ipu))if diffLen >= k: print(0)else: if k > ipuLen: print('impossible') else: print(k - diffLen)
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