Harmonic Number (II) LightOJ - 1245

最新推荐文章于 2019-07-11 12:39:20 发布

原创最新推荐文章于 2019-07-11 12:39:20 发布 · 203 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种解决1234-HarmonicNumber问题的方法，通过优化整数除法来提高计算效率。输入包含多个测试用例，每个用例给出一个整数n，需要计算H(n)的值。

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.

#include <cstdio> 
#include<iostream> 
#include <cmath> 
using namespace std;
long long n,m,res;
int main(){
    int t,kase=1,r,l;
    scanf("%d",&t);
    while(t--){
        cin>>n;
        res=0;
        m=int(sqrt(n));
        for(int i=1;i<=m;i++){
            if(n/i!=i)
                res+=n/i;
            l=n/i;
            r=n/(i+1);
            res+=i*(l-r);
        }
        printf("Case %d: %lld\n",kase++,res);
    }
    return 0;
}