ICPC Central Europe Regional Contest 2019 E. Zeldain Garden（除法分塊，反函數求積分）

Boris is the chief executive officer of Rock Anywhere Transport (RAT) company which specializes in supporting music industry. In particular, they provide discount transport for manypopular rock bands. This time Boris has to move a large collection of quality Mexican concertloudspeakers from the port on the North Sea to the far inland capital. As the collection isexpected to be big, Boris has to organize a number of lorries to assure smooth transport. Themultitude of lorries carrying the cargo through the country is called a convoy

Boris wants to transport the whole collection in one go by a single convoy and without leavingeven a single loudspeaker behind. Strict E.U. regulations demand that in the case of largetransport of audio technology, all lorries in the convoy must carry exactly the same number ofpieces of the equipment.

To meet all the regulations, Boris would like to do some planning in advance, despite the fact thathe does not yet know the exact number of loudspeakers, which has a very significant influenceon the choices of the number and the size of the lorries in the convoy. To examine variousscenarios, for each possible collection size, Boris calculates the so-called “variability”, which isthe number of different convoys that may be created for that collection size without violatingthe regulations. Two convoys are different if they consist of a different number of lorries.

For instance, the variability of the collection of 6 loudspeakers is 4, because they may be evenlydivided into 1, 2, 3, or 6 lorries.

Input Specification

The input contains one text line with two integers N, M (1 \leq N \leq M \leq 10^{12})N,M(1≤N≤M≤10
12
), the minimumand the maximum number of loudspeakers in the collection.

Output Specification

Print a single integer, the sum of variabilities of all possible collection sizes betweenNN and MM,inclusive.

樣例輸入1複製
2 5
樣例輸出1複製
9
樣例輸入2複製
12 12
樣例輸出2複製
6
樣例輸入3複製
555 666
樣例輸出3複製
852

題意；
求所有n~m間每個數的約數數目和。

思路：
求前$n$個數的約數數目和，就是求$∑(n/i)$
這個過程可以用除法分塊加速（對於所有 n/i 值相同的部分進行分塊）。

貌似還看到了一種玄學的解法，反函數$f(x) = n/x$面積求和來求解。
這個函數按照 ($\sqrt{x},\sqrt{x})$對稱，所以結果爲sqrt(x)前的所有∑n/i乘以2，再減去sqrt(x) * sqrt(x)，這個過程可以畫圖看出來。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long ll;

ll solve(ll n) {
    ll ans = 0;
    for(ll i = 1,j = 1;i <= n;) {
        j = n / (n / i);
        ans += (j - i + 1) * (n / i);
        i = j + 1;
    }
    return ans;
}

ll solve2(ll n) {
    ll ans = 0;
    ll t = sqrt(n);
    for(ll i = 1;i <= t;i++) {
        ans += n / i;
    }
    return ans * 2 - t * t;
}

int main() {
    ll n,m;scanf("%lld%lld",&m,&n);
    printf("%lld\n",solve2(n) - solve2(m - 1));
    return 0;
}