LightOJ 1245 数学

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 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

11

1

2

3

4

5

6

7

8

9

10

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

思路:类似分块的思想最多有2sqrt(n)个数;

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define esp 0.00000000001
#define pi 4*atan(1)
const int N=1e3+10,M=1e7+10,inf=1e9+10,mod=1e9+7;
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        ll x;
        scanf("%lld",&x);
        ll ans=0;
        for(ll i=1;i*i<=x;i++)
        {
            if(i!=x/i)
            ans+=x/i;
            ll len=(x/i)-x/(i+1);
            ans+=i*len;
        }
        printf("Case %d: %lld
",cas++,ans);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/jhz033/p/5785270.html