題目連結

https://lydsy.com/JudgeOnline/problem.php?id=3529

題解

題目要求
$\underset{i=1}{\overset{n}{\sum }}$

∑ j = 1 m σ 0

( gcd ⁡ ( i , j ) ) [ σ 0 ( gcd ⁡ ( i , j ) ) ≤ a ] \sum_{i=1}^n\sum_{j=1}^m \sigma_0(\gcd(i,j))[\sigma_0(\gcd(i,j))\leq a]
稍微轉化一下就是
$\sum_{T=1}^{\min} \lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\sum_{k|T}\mu(\frac{T}{k})\sigma_0(k)[\sigma_0(k)\leq a]$
設
$f(T)=\sum_{k|T}\mu(\frac{T}{k})\sigma_0(k)[\sigma_0(k)\leq a]$
考慮對於每一個 $\sigma_0(k)=a$，更新其在對應位置上的 $f$，可以預處理出 $\sigma_0$，然後以權值為關鍵字sort一邊，離線處理詢問，將詢問以 $a$為關鍵字sort一遍，用樹狀陣列維護每個位置的 $f$。

這裡有一個小技巧，由於對 $2^{31}$取模，因此可以利用自然溢位，用unsigned來記錄答案，最後取模即可。

時間複雜度 $O(T\sqrt{n}\log n+n\log^2 n)$。

程式碼

#include <cstdio>
#include <algorithm>
 
int read()
{
  int x=0,f=1;
  char ch=getchar();
  while((ch<'0')||(ch>'9'))
    {
      if(ch=='-')
        {
          f=-f;
        }
      ch=getchar();
    }
  while((ch>='0')&&(ch<='9'))
    {
      x=x*10+ch-'0';
      ch=getchar();
    }
  return x*f;
}
 
const int maxn=100000;
const int maxq=20000;
const unsigned mod=1<<31;
 
namespace bit
{
  unsigned val[maxn+10];
 
  int lowbit(int x)
  {
    return x&(-x);
  }
 
  int add(int pos,unsigned v)
  {
    while(pos<=maxn)
      {
        val[pos]+=v;
        pos+=lowbit(pos);
      }
    return 0;
  }
 
  unsigned getsum(int pos)
  {
    unsigned res=0;
    while(pos)
      {
        res+=val[pos];
        pos-=lowbit(pos);
      }
    return res;
  }
}
 
struct data
{
  int id;
  unsigned val;
 
  bool operator <(const data &other) const
  {
    return val<other.val;
  }
};
 
int p[maxn+10],prime[maxn+10],cnt,low[maxn+10],num[maxn+10];
unsigned mu[maxn+10];
data d[maxn+10];
 
unsigned F(int x,int y)
{
  unsigned res=1;
  for(int i=1; i<=y+1; ++i)
    {
      res*=x;
    }
  return (res-1)/(x-1);
}
 
int getprime()
{
  for(int i=1; i<=maxn; ++i)
    {
      d[i].id=i;
    }
  p[1]=mu[1]=num[1]=d[1].val=1;
  low[1]=0;
  for(int i=2; i<=maxn; ++i)
    {
      if(!p[i])
        {
          prime[++cnt]=i;
          low[i]=num[i]=1;
          mu[i]=-1;
          d[i].val=i+1;
        }
      for(int j=1; (j<=cnt)&&(i*prime[j]<=maxn); ++j)
        {
          int x=i*prime[j];
          p[x]=1;
          if(i%prime[j]==0)
            {
              low[x]=low[i]+1;
              num[x]=num[i];
              mu[x]=0;
              d[x].val=d[num[x]].val*F(prime[j],low[x]);
              break;
            }
          low[x]=1;
          num[x]=i;
          mu[x]=-mu[i];
          d[x].val=d[i].val*(prime[j]+1);
        }
    }
  std::sort(d+1,d+maxn+1);
  return 0;
}
 
struct query
{
  int n,m,v,id;
 
  bool operator <(const query &other) const
  {
    return v<other.v;
  }
};
 
int T;
query q[maxq+10];
unsigned ans[maxn+10];
 
int main()
{
  getprime();
  T=read();
  for(int i=1; i<=T; ++i)
    {
      q[i].n=read();
      q[i].m=read();
      q[i].v=read();
      q[i].id=i;
    }
  std::sort(q+1,q+T+1);
  for(int i=1,j=1; i<=T; ++i)
    {
      while((j<=maxn)&&(d[j].val<=(unsigned)q[i].v))
        {
          for(int k=1; k<=maxn/d[j].id; ++k)
            {
              bit::add(d[j].id*k,mu[k]*d[j].val);
            }
          ++j;
        }
      for(int l=1,r; l<=std::min(q[i].n,q[i].m); l=r+1)
        {
          r=std::min(q[i].n/(q[i].n/l),q[i].m/(q[i].m/l));
          ans[q[i].id]+=(bit::getsum(r)-bit::getsum(l-1))*(q[i].n/l)*(q[i].m/l);
        }
    }
  for(int i=1; i<=T; ++i)
    {
      printf("%u\n",ans[i]&(mod-1));
    }
  return 0;
}