嘟嘟嘟


学完数论分块，觉得这题不难啊。（难道是我变强了？）


推式子就行。

\[\begin{align*} G(n, k) &= \sum_{i = 1} ^ {n} k \ \ mod \ \ i \\ &= \sum_{i = 1} ^ {n} k - \left \lfloor \frac{k}{i} \right \rfloor * i \\ &= n * k - \sum_{i = 1} ^ {n} \left \lfloor \frac{k}{i} \right \rfloor * i \end{align*}\]

然后发现后面的\(\sum_{i = 1} ^ {n} \left \lfloor \frac{k}{i} \right \rfloor * i\)可以用数论分块来做。
不过需要注意的是\(n,k\)大小关系不一定，所以块的右端点应该是\(min(n, \frac{k}{ \left \lfloor \frac{k}{l} \right \rfloor})\)。
以及循环终止条件为\(l \leqslant min(n, k)\)。
然后块内等差数列求和即可。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
//const int maxn = ;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

ll n, k, ans = 0;

int main()
{
  n = read(); k = read();
  ans = n * k;
  for(ll l = 1, r; l <= min(n, k); l = r + 1)
    {
      r = min(n, k / (k / l));
      ans -= (k / l) * (r - l + 1) * (l + r) / 2;
    }
  write(ans), enter;
  return 0;
}