算是一个找规律的题目吧. 枚举前sqrt(n)个数,数i出现的次数为n/i-n/(i+1),对答案的贡献为(n/i-n/(i+1))*i. 对于sqrt后边的数,可以直接由n/i获得,并且一定只出现一次. (数学果然博大精深~~~~) code: #include<bits/stdc++.h> using namespace std; typedef long long ll; void solve(ll time){ ll n; cin>>n; ll ans=; ll c=sqr…

题意: 求前n项的n/i  的和 只取整数部分 暴力肯定超时...然后 ...现在的人真聪明...我真蠢 觉得还是别人的题意比较清晰 比如n=100的话,i=4时n/i等于25,i=5时n/i等于20,于是在大于20到小于等于25内的5个数字j都有n/j等于4,然后ans+=4*5 所以我们可以在小于等于根号n的范围内枚举i,ans+=n/i,然后ans+=(n/(i)-n/(i+1))*i,这样分段加起来 但是又重复的部分.. 即 令m = sqrt(n), 如果n / m == m 则n /…

/** 题目:G - Harmonic Number (II) 链接:https://vjudge.net/contest/154246#problem/G 题意:给定一个数n,求n除以1~n这n个数的和.n达到2^31 - 1; 思路: 首先我们观察一下数据范围,2^31次方有点大,暴力会超时,所以我们看看有没有啥规律,假设 tmp 是 n/i 的值,当n == 10的时候(取具体值) 当 tmp = 1 时,个数 是10/1 - 10/2 == 5个 当 tmp = 2 时,个数 是10/2…

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…

1245 - Harmonic Number (II)   PDF (English) Statistics Forum Time Limit: 3 second(s) Memory Limit: 32 MB 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 =…

题目链接:https://vjudge.net/problem/LightOJ-1245 1245 - Harmonic Number (II)    PDF (English) Statistics Forum Time Limit: 3 second(s) Memory Limit: 32 MB I was trying to solve problem '1234 - Harmonic Number', I wrote the following code long long H( int…

Harmonic Number (II)   PDF (English) Statistics Forum Time Limit: 3 second(s) Memory Limit: 32 MB 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 <= …

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…

题解:隔一段数字存一个答案,在查询时,只要找到距离n最近而且小于n的存答案值,再把剩余的暴力跑一遍就可以. #include <bits/stdc++.h> using namespace std; const int N = 1e8 + 10; const int M = 2e6 + 10; double a[M]; void Init() { a[0] = 0.0; double ans = 1; for( int i = 2; i < N; i ++) { ans += 1.0 /…

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 was that I was using the integer…