嘟嘟嘟


感覺這幾道數論題都差不多，但這到明顯是前幾道的升級版。
推了一大頓只能得60分，不得不看題解。
各位看這老哥的題解吧
我就是推到他用\(T\)換掉\(kd\)之前，然後列舉\(T\)的。這個轉換確實想不出來啊。
還有最後一句，最終的式子
\[\sum_{T = 1} ^ {n} \lfloor \frac{n}{T} \rfloor \lfloor \frac{m}{T} \rfloor * \sum_{k | T} \mu(\frac{T}{k}) (k \in prime)\]
他把後面的那個sum預處理了。令\(f(T) = \sum_{k | T} \mu(\frac{T}{k}) (k \in prime)\)

#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 = 1e7 + 5;
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');
}

int v[maxn], prm[maxn], mu[maxn];
ll f[maxn], sum[maxn];
void init()
{
  mu[1] = 1;
  for(int i = 2; i < maxn; ++i)
    {
      if(!v[i]) v[i] = i, prm[++prm[0]] = i, mu[i] = -1;
      for(int j = 1; j <= prm[0] && i * prm[j] < maxn; ++j)
    {
      v[i * prm[j]] = prm[j];
      if(i % prm[j] == 0) {mu[i * prm[j]] = 0; break;}
      else mu[i * prm[j]] = -mu[i];
    }
    }
  for(int i = 1; i <= prm[0]; ++i)
    for(int j = 1; prm[i] * j < maxn; ++j)
      f[prm[i] * j] += mu[j];
  for(int i = 1; i < maxn; ++i) sum[i] = sum[i - 1] + f[i];
}

ll solve(int n, int m)
{
  int Min = min(n, m);
  ll ret = 0;
  for(int l = 1, r; l <= Min; l = r + 1)
    {
      r = min(n / (n / l), m / (m / l));
      ret += (sum[r] - sum[l - 1]) * (n / l) * (m / l);
    }
  return ret;
}

int main()
{
  init();
  int T = read();
  while(T--)
    {
      ll n = read(), m = read();
      write(solve(n, m)), enter;
    }
  return 0;
}