摘要：
CodeChef Expected Repetitions 记 \(\operatorname{power}(S)\) 为字符串 \(S\) 所有子串的幂值和，显然答案即为 \(\dfrac{2\operatorname{power}(S)}{N(N+1)}\)。考虑如何计算 \(\operat 阅读全文
posted @ 2021-12-28 16:10
cutx64
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