YBT高效进阶 6.2.5 余数之和
思路
( n , k ) (n,k) (n,k)
= ∑ i = 1 n k m o d i =\sum_{i=1}^nk\bmod i =∑i=1nkmodi
= ∑ i = 1 n k − ⌊ k i ⌋ =\sum_{i=1}^nk-\left\lfloor \dfrac ki \right\rfloor =∑i=1nk−⌊ik⌋
= n ∗ k − ∑ i = 1 n ⌊ k i ⌋ =n*k-\sum_{i=1}^n\left\lfloor \dfrac ki \right\rfloor =n∗k−∑i=1n⌊ik⌋
分块
i ∈ [ x , ⌊ k ⌊ k x ⌋ ⌋ ] 时 ⌊ k i ⌋ 的 值 相 等 i\in[x,\left\lfloor \dfrac k{\left\lfloor\dfrac kx\right\rfloor} \right\rfloor]时\left\lfloor \dfrac ki \right\rfloor的值相等 i∈[x,⎣⎢⎢⎢⎢⎢⌊xk⌋k⎦⎥⎥⎥⎥⎥]时⌊ik⌋的值相等
再用等差数列
( x + ⌊ k ⌊ k x ⌋ ⌋ ) / 2 (x+\left\lfloor \dfrac k{\left\lfloor\dfrac kx\right\rfloor} \right\rfloor)/2 (x+⎣⎢⎢⎢⎢⎢⌊xk⌋k⎦⎥⎥⎥⎥⎥)/2
n ∗ k − ∑ i = 1 n ⌊ k i ⌋ n*k-\sum_{i=1}^n\left\lfloor \dfrac ki \right\rfloor n∗k−∑i=1n⌊ik⌋
= n ∗ k − ∑ ( x + ⌊ k ⌊ k x ⌋ ⌋ ) ∗ ( ⌊ k ⌊ k x ⌋ ⌋ − x + 1 ) / 2 ∗ ⌊ k x ⌋ =n*k-\sum(x+\left\lfloor \dfrac k{\left\lfloor\dfrac kx\right\rfloor} \right\rfloor)*(\left\lfloor \dfrac k{\left\lfloor\dfrac kx\right\rfloor} \right\rfloor-x+1)/2*\left\lfloor \dfrac kx \right\rfloor =n∗k−∑(x+⎣⎢⎢⎢⎢⎢⌊xk⌋k⎦⎥⎥⎥⎥⎥)∗(⎣⎢⎢⎢⎢⎢⌊xk⌋k⎦⎥⎥⎥⎥⎥−x+1)/2∗⌊xk⌋
CODE
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
long long ans,n,k,i,x;
scanf("%lld%lld",&n,&k);
for(ans=n*k,i=1;i<=n;i=x+1)x=(k/i)?min(n,k/(k/i)):n,ans-=k/i*(i+x)*(x-i+1)/2;
printf("%lld",ans);
return 0;
}