light_oj 1245 求[n/i]的前n项和

最新推荐文章于 2022-04-23 16:15:30 发布

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最新推荐文章于 2022-04-23 16:15:30 发布

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原文链接：http://www.cnblogs.com/--560/p/4550678.html

本文针对light_oj1245问题，提出了一种高效的算法来计算[n/i]的前n项和。该算法通过将计算过程分为两部分——i小于等于sqrt(n)和i大于sqrt(n)——来优化计算效率，并给出了完整的C++实现代码。

light_oj 1245 求[n/i]的前n项和

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 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 divisions only. However, you are given n, you have to find H(n) as in my code.

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n < 2³¹).

Output

For each case, print the case number and H(n) calculated by the code.

Sample Input

2147483647

Sample Output

Case 1: 1

Case 2: 3

Case 3: 5

Case 4: 8

Case 5: 10

Case 6: 14

Case 7: 16

Case 8: 20

Case 9: 23

Case 10: 27

Case 11: 46475828386

题意：求[n/i]的前n项和

思路：在i=1到sqrt(n)时，[n/i]是不重复的，直接计算；在i>sqrt(n)时，反着数，从n/n，n/(n-1)，...，n/(n/2+1)，每个值为1，总共(n-n/2)个，以此类推，值为i的个数为i*(n/i-n/(i-1))，i应该是小于等于sqrt(n)的，而当[sqrt(n)]==n/[sqrt(n)]时，[sqrt(n)]被计算了两次，特判即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<string>
#include<math.h>
#include<cctype>

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=(1<<29);
const double EPS=0.0000000001;
const double Pi=acos(-1.0);

ll n;

ll f(ll n)
{
    ll res=0;
    for(ll i=1;i<=sqrt(n);i++) res+=n/i;
    for(ll i=1;i<=sqrt(n);i++) res+=i*(n/i-n/(i+1));
    if((ll)sqrt(n)==n/(ll)sqrt(n)) res-=(ll)sqrt(n);
    return res;
}

int main()
{
    int T;cin>>T;
    int tag=1;
    while(T--){
        cin>>n;
        printf("Case %d: %lld\n",tag++,f(n));
    }
    return 0;
}