http://lightoj.com/volume_showproblem.php?problem=1245 G - Harmonic Number (II) Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1245 Description I was trying to solve problem '1234 - Harmonic…
Harmonic Number In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers: In this problem, you are given n, you have to find Hn. Input Input starts with an integer T (≤ 10000), denoting the number of test c…
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=98634#problem/B(acm14) Description I was trying to solve problem '1234 - Harmonic Number', I wrote the following code long long H( int n ) { long long res = 0; for( int i = 1; i <= n; i++ ) …
题目大意:对下列代码进行优化 long long H( int n ) { long long res = 0; for( int i = 1; i <= n; i++ ) res = res + n / i; return res;} 题目思路:为了避免超时,要想办法进行优化 以9为例: 9/1 = 9 9/2 = 4 9/3 = 3 9/4 = 2 9/5 = 1 9/6 = 1 9/7 = 1 9/8 = 1 9/9 = 1 拿1来看,同为1的区间长度为:9…
打表或者画个图可以看出i>根号n时每个i的贡献值相差很小,可以利用公式优化(函数C) 但是注意不能一整段使用公式,否则复杂度还是会劣化到O(n)(显然对gongxian只能逐步递减) 网上看了不少代码,但是都没有对贡献值边界问题给定明确的判断 所以还是加多一个while循环确定贡献值的开端是前面的n/i没有的 #include<bits/stdc++.h> using namespace std; const int maxn = 1e5+11; typedef long long ll…
Harmonic Number (II) Description I was trying to solve problem '1234 - Harmonic Number', I wrote the following code long long H( int n ) { long long res = 0; for( int i = 1; i <= n; i++ ) res = res + n / i; return res; } Yes, my e…
链接: https://vjudge.net/problem/LightOJ-1245 题意: I was trying to solve problem '1234 - Harmonic Number', I wrote the following code long long H( int n ) { long long res = 0; for( int i = 1; i <= n; i++ ) res = res + n / i; return res; } Yes, my error…
ID Origin Title 111 / 423 Problem A LightOJ 1370 Bi-shoe and Phi-shoe 21 / 74 Problem B LightOJ 1356 Prime Independence 61 / 332 Problem C LightOJ 1341 Aladdin and the Flying Carpet 54 / 82 Problem D LightOJ 1336 Sigma Function 66 /…
Pairs Forming LCM (LightOJ - 1236)[简单数论][质因数分解][算术基本定理](未完成) 标签: 入门讲座题解 数论 题目描述 Find the result of the following code: long long pairsFormLCM( int n ) { long long res = 0; for( int i = 1; i <= n; i++ ) for( int j = i; j <= n; j++ ) if( lcm(i, j) ==…
Help Hanzo (LightOJ - 1197) [简单数论][筛区间质数] 标签: 入门讲座题解 数论 题目描述 Amakusa, the evil spiritual leader has captured the beautiful princess Nakururu. The reason behind this is he had a little problem with Hanzo Hattori, the best ninja and the love of Nakurur…
Aladdin and the Flying Carpet (LightOJ - 1341)[简单数论][算术基本定理][分解质因数](未完成) 标签:入门讲座题解 数论 题目描述 It's said that Aladdin had to solve seven mysteries before getting the Magical Lamp which summons a powerful Genie. Here we are concerned about the first myste…
Sigma Function (LightOJ - 1336)[简单数论][算术基本定理][思维] 标签: 入门讲座题解 数论 题目描述 Sigma function is an interesting function in Number Theory. It is denoted by the Greek letter Sigma (σ). This function actually denotes the sum of all divisors of a number. For exam…
Goldbach`s Conjecture(LightOJ - 1259)[简单数论][筛法] 标签: 入门讲座题解 数论 题目描述 Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states: Every even integer, greater than 2, can be expressed as the sum of…
