LightOJ1245 Harmonic Number (II) —— 规律

题目链接：https://vjudge.net/problem/LightOJ-1245

1245 - Harmonic Number (II)

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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 <= 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	Output for Sample Input
11 1 2 3 4 5 6 7 8 9 10 2147483647	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：

对于一个数n，求出 sigma(n/k)， 1<=k<=n。

题解：

1.由于n<=2^31，直接暴力不不可能的。

2.手写一下可发现一个规律， 当 n/i = k时， i的范围为 (n/(k+1), n/k]，即有n/k- n/(k+1) 个 i 使得n/i = k。

3.有了上述结论之后，就可以降低暴力程度了，设 m = sqrt(n)。先从1枚举到m，求出n/i的和，这样 [1,m] 这个区间的求值就解决了，还剩 [m+1, n] 这个区间：从m递减枚举到1，求出k*(n/k - n/(k+1))的和，这样就求出了[ n/m, n]的值，当 n/m==m时，即开始下标为m，则区间为[m,n]，所以在m的位置重复计算了，需要减去m；当n/m!=m，即n/m>m时，其实下标为m+1，所以区间为 [m+1,n]。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int MAXM = 1e5+10;
const int MAXN = 1e6+10;

int main()
{
    int T, n, kase = 0;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        int m = sqrt(n);

        LL ans = 0;
        for(int i = 1; i<=m; i++) ans += n/i;
        for(int i = 1; i<=m; i++) ans += 1LL*i*(n/i - n/(i+1));
        if(n/m==m) ans -= m;
        printf("Case %d: %lld
", ++kase, ans);
    }
}