http://lightoj.com/volume_showproblem.php?problem=1234 Harmonic Number Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1234 Description In mathematics, the nth harmonic number is the sum of th…

Harmonic Number Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1234 Description In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers: In this p…

D - Harmonic Number Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1234 Description In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers: In th…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1234 Sample Input Sample Output Case : Case : 1.5 Case : 1.8333333333 Case : 2.0833333333 Case : 2.2833333333 Case : 2.450 Case : 2.5928571429 Case : 2.7178571429 Case : 2.8289682540 Case : 18.89…

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://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…

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…

题意: 求调和级数,但n很大啦.. 解析: 分段打表  每间隔50存储一个数,在计算时  只需要找到离输入的n最近的那个数 以它为起点 开始计算即可 emm...补充一下调和级数的运算公式   r为常数,r=0.57721566490153286060651209(r就是欧拉常数). 看一下这位的博客:https://www.cnblogs.com/weiyuan/p/5737273.html #include <iostream> #include <cstdio> #inclu…

题目给的时间限制是3s,所以可以直接暴力来做,注意n的取值范围是1e8,如果开一个1e8的数组会RE.分段打表,可以每100个数记录一次,然后对每次询问先找到它所在的区间,然后在暴力往后找.(学到了~~) #include<bits/stdc++.h> using namespace std; ; double mp[N]; void inint(){ ; ;i<=;i++){ ans+=(double)1.0/i; ==) mp[i/]=ans; } } int main(){ ini…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1234 给你一个数n,让你求 这个要是直接算的话肯定TLE,要是用1e8的数组预处理存储答案也一定MLE. 所以我用1e6的数组存储每100个数的答案,然后每次给你n的时候顶多算99次. #include <iostream> #include <cstdio> #include <cstring> using namespace std; const int…