【整除分块】 余数之和

传送门

题意

给定正整数(n，k)，计算((k ; mod; 1)+(k; mod; 2)+(k; mod; 3)+dots +(k; mod; n))的值,

数据范围

(1leq n,k leq 10^{9})

题解

(k; mod ; i = k- lfloor frac{k}{i} floor imes i)，问题转化为计算(n imes k - sum_{i=1}^{n} lfloor frac{k}{i} floor imes i)
证明(sum_{i=1}^{n}leftlfloorfrac{k}{i} ight floor imes i)最多只有(2 sqrt{k})个值，

• 当(i leq sqrt{k})的时候，最多有(sqrt{k})个不同的取值
• 当(i>sqrt{k})的时候，(leftlfloorfrac{k}{i} ight floor < sqrt{k}) , 故最多有(sqrt{k})个不同的取值

易知(leftlfloorfrac{k}{x} ight floor)相同的必定是连续的，若某一个相等区间(x)是下界，上界是(leftlfloorfrac{k}{leftlfloorfrac{k}{x} ight floor} ight floor)
证明：
令 (l = x)，(r = leftlfloorfrac{k}{leftlfloorfrac{k}{x} ight floor} ight floor), (leftlfloorfrac{k}{x} ight floor)显然随着(x)增加减少，

[ecauseleftlfloorfrac{k}{x} ight floorleq frac{k}{x}， herefore r geqslantleftlfloorfrac{k}{left(frac{k}{x} ight)} ight floor=x ]

[ hereforeleftlfloorfrac{k}{r} ight floor leqleftlfloorfrac{k}{x} ight floor ]

[egin{array}{l}ecause r leq frac{k}{leftlfloorfrac{k}{x} floor ight.} , herefore leftlfloorfrac{k}{r} ight floor geqslantleftlfloorfrac{k}{frac{k}{leftlfloorfrac{k}{x} ight floor}} ight floorend{array} ]

[ herefore leftlfloorfrac{k}{r} ight floor geqslant x ]

综上可以得到

[leftlfloor frac{k}{r} ight floor=leftlfloorfrac{k}{x} ight floor ]

[ecause i inleft[x,leftlfloorfrac{k}{leftlfloorfrac{k}{x} ight floor} ight floor ight]区间内的数,lfloorfrac{k}{i} floor的值都相等 ]

时间复杂度：(O(sqrt{k}))

Code

#include<bits/stdc++.h>
using namespace std;
#define ll long long

ll ans;

int main()
{
    ll n,k,l,r;
    cin>>n>>k;
    ans=n*k;
    for(l = 1;l<=n;l=r+1)
    {
        if(k/l == 0) break;//0对答案无贡献

        r=min(k/(k/l),n);
        ans-=1ll * (k/l) *(l+r)*(r-l+1)/2;//等差数列求和你n*(a1+an)/2
    }
    cout<<ans<<endl;
}