SPOJ AMR11E Distinct Primes 基础数论

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SPOJ AMR11E Distinct Primes 基础数论

Arithmancy is Draco Malfoy's favorite subject, but what spoils it for him is that Hermione Granger is in his class, and she is better than him at it. Prime numbers are of mystical importance in Arithmancy, and Lucky Numbers even more so. Lucky Number…

SPOJ 10232. Distinct Primes

Arithmancy is Draco Malfoy's favorite subject, but what spoils it for him is that Hermione Granger is in his class, and she is better than him at it. Prime numbers are of mystical importance in Arithmancy, and Lucky Numbers even more so. Lucky Numbe…

（Problem 47）Distinct primes factors

The first two consecutive numbers to have two distinct prime factors are: 14 = 2 7 15 = 3 5 The first three consecutive numbers to have three distinct prime factors are: 644 = 2² 7 23 645 = 3 5 43 646 = 2 17 19. Find the first four consecutiv…

【SPOJ】Distinct Substrings（后缀自动机）

[SPOJ]Distinct Substrings(后缀自动机) 题面 Vjudge 题意:求一个串的不同子串的数量题解对于这个串构建后缀自动机之后我们知道每个串出现的次数就是\(right/endpos\)集合的大小但是实际上我们没有任何必要减去不合法的数量我们只需要累加每个节点表示的合法子串的数量即可这个值等于\(longest-shortest+1=longest-parent.longest\) #include<iostream> #include<cstdio&g…

【SPOJ】Distinct Substrings/New Distinct Substrings（后缀数组）

[SPOJ]Distinct Substrings/New Distinct Substrings(后缀数组) 题面 Vjudge1 Vjudge2 题解要求的是串的不同的子串个数两道一模一样的题目其实很容易: 总方案-不合法方案数对于串进行后缀排序后不合法方案数=相邻两个串的不合法方案数的和也就是\(height\)的和所以\[ans=\frac{n(n+1)}{2}-\sum_{i=1}^{len}height[i]\] #include<iostream> #include…

LightOJ1214 Large Division 基础数论+同余定理

Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c. Input Input starts with an integer T (≤ 525), denot…