技術標籤：牛客演算法數學c++c語言

題目連結：https://ac.nowcoder.com/acm/contest/7817/E

題目描述
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 many popular rock bands. This time Boris has to move a large collection of quality Mexican concert loudspeakers from the port on the North Sea to the far inland capital. As the collection is expected to be big, Boris has to organize a number of lorries to assure smooth transport. The multitude 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 leaving even a single loudspeaker behind. Strict E.U. regulations demand that in the case of large transport of audio technology, all lorries in the convoy must carry exactly the same number of pieces of the equipment.
To meet all the regulations, Boris would like to do some planning in advance, despite the fact that he does not yet know the exact number of loudspeakers, which has a very significant influence on the choices of the number and the size of the lorries in the convoy. To examine various scenarios, for each possible collection size, Boris calculates the so-called “variability”, which is the number of different convoys that may be created for that collection size without violating the 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 evenly divided into 1, 2, 3, or 6 lorries.

輸入描述:
The input contains one text line with two integers N, M (1 ≤ N ≤ M ≤ 10^12 ), the minimum and the maximum number of loudspeakers in the collection.

輸出描述:
Print a single integer, the sum of variabilities of all possible collection sizes between N and M,inclusive.

示例1

輸入

2 5

輸出

9

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分析

數論中有個很牛的結論：
n之內的所有因數個數的計算方法為：n / 1 + n / 2 + n / 3 … n / n ；

還有個問題， n 可以到 10^12 ，不能直接線性。
那麼例如當我們列舉 n = 12 時，發現 12 / (7, 8, 9, 10, 11, 12) 都為 1 ，我們可以只計算一次把他們全部計算出來；
或者發現，y = n / x 這個函式關於 y = x 對稱，對稱為點（√n，√n），因此由函式影象面積對稱可以知道，只需求出前 √n 資料範圍內的面積 ans ， ans * 2 - √n * √n 即為答案。

程式碼

# include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL X(LL n)
{
    LL l, r;
    LL ans = 0;
    for( l = 1; l <= n; l = r+1)
    {
        r = n/(n/l);
        ans += n/l *(r - l + 1);
    }
    return ans;
    /*
    for(l=1; l<=t; l++)
            ans += n/l;
    ans = ans*2-t*t;
    return ans;
    */
}
int main(){
    int t;
        LL x,y;
        cin>> x>>y;
        printf("%lld\n",X(y)-X(x-1));