Counting trailing 0s of n! It is not very hard to figure out how to count it - simply count how many 5s. Since even numbers are always more than 5s, we don't have to consider even numbers. But counting 5, is tricky. Naive method is quite slow - I got…
参考高性能javascript for in 循环 使用它可以遍历对象的属性名,但是每次的操作都会搜索实例或者原型的属性 导致使用for in 进行遍历会产生更多的开销 书中提到不要使用for in 遍历数组 1 首先for in 会查找原型链上的属性 var arr = [1,2,3]; Array.prototype.a = "test"; for(var i in arr) { console.log(i); console.log(typeof i); }//在这里例子中会发…
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Factorial numbers are getting big very soon, you'll have to compute the number of divisors of such highly composite numbers. Input The first line contains an integer T, the number of test cases.On the next T lines, you will be given two integers N an…
传送门 双倍经验(弱化版本) 考虑求出来heightheightheight数组之后用增量法. 也就是考虑每增加一个heightheightheight对答案产生的贡献. 算出来是∑∣S∣−heighti+1−sai\sum|S|-height_i+1-sa_i∑∣S∣−heighti+1−sai 代码: #include<bits/stdc++.h> #define ri register int using namespace std; const int N=5e4+5; int n…
[codeforces 516]A. Drazil and Factorial 试题描述 Drazil is playing a math game with Varda. Let's define for positive integer x as a product of factorials of its digits. For example, . First, they choose a decimal number a consisting of n digits that con…
Coding a Dijkstra is not hard. %70 of my time spent on tackling TLE, as my last post. Dijkstra works in BFS manner, but at each step, it picks the shortest child greedily and then relax all other neighbors. Data Structure: since there could be up to…
My first idea was Sieve of Eratosthenes, too. But obviously my coding was not optimal and it exceeded 6s time limit. Then I googled it. Sounds like Miller-Rabin Testing is a much more serious solution. http://www.cnblogs.com/feature/articles/1824667.…
