P6810 「MCOI-02」Convex Hull 凸包

「MCOI-02」Convex Hull 凸包

题意即求：

$$\sum _{i=1}^n\sum _{j=1}^m{\sigma }_0(i){\sigma }_0(j){\sigma }_0(gcd(i,j))(\mathrm{mod}p)$$

其中 $T=1$，$n,m\le 2\times {10}^6$，$p\le {10}^9$。

被 long long 制裁了，诡异的常数差距。

剩下就是一道水题。

推导一下结束吧：

$$\begin{array}{rl}& \sum _{i=1}^n\sum _{j=1}^m{\sigma }_0(i){\sigma }_0(j){\sigma }_0(gcd(i,j))\\ =& \sum _{d=1}\sum _{i=1}^n\sum _{j=1}^m{\sigma }_0(i){\sigma }_0(j){\sigma }_0(d)[gcd(i,j)=d]\\ =& \sum _{d=1}{\sigma }_0(d)\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor m/d\rfloor }{\sigma }_0(id){\sigma }_0(jd)[gcd(i,j)=1]\\ =& \sum _{d=1}{\sigma }_0(d)\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor m/d\rfloor }{\sigma }_0(id){\sigma }_0(jd)\sum _{k\mid i,k\mid j}\mu (k)\\ =& \sum _{d=1}{\sigma }_0(d)\sum _{k=1}\mu (k)\sum _{i=1}^{\lfloor n/kd\rfloor }{\sigma }_0(ikd)\sum _{j=1}^{\lfloor m/kd\rfloor }{\sigma }_0(jkd)\\ =& \sum _{T=1}\sum _{d\mid T}{\sigma }_0(d)\mu \left({\displaystyle \frac{T}{d}}\right)\sum _{i=1}^{\lfloor n/T\rfloor }{\sigma }_0(iT)\sum _{j=1}^{\lfloor m/T\rfloor }{\sigma }_0(jT)\end{array}$$

然后随便做就好了。

时间复杂度 $O(n\ln n)$。

代码：

#include<iostream>
#include<cstdio>
#define ll int
using namespace std;
namespace Ehnaev{
  inline ll read() {
    ll ret=0,f=1;char ch=getchar();
    while(ch<48||ch>57) {if(ch==45) f=-f;ch=getchar();}
    while(ch>=48&&ch<=57) {ret=(ret<<3)+(ret<<1)+ch-48;ch=getchar();}
    return ret*f;
  }
  inline void write(ll x) {
    static char buf[22];static ll len=-1;
    if(x>=0) {do{buf[++len]=x%10+48;x/=10;}while(x);}
    else {putchar(45);do{buf[++len]=-(x%10)+48;x/=10;}while(x);}
    while(len>=0) putchar(buf[len--]);
  }
}using Ehnaev::read;using Ehnaev::write;

const ll N=2e6;

ll n,m,p,cnt,ans;
ll sgm0[N+5],prime[N+5],f[N+5],g[N+5];
bool ff[N+5];

inline void Init() {
  for(ll i=1;i<=n;i++) {
    for(ll j=i;j<=n;j+=i) {
      sgm0[j]=(sgm0[j]+1)%p;
    }
  }
  for(ll i=1;i<=n;i++) {
    for(ll j=i,cn=1;j<=n;j+=i,cn++) {
      f[i]=(f[i]+sgm0[j])%p;
    }
  }
  for(ll i=1;i<=m;i++) {
    for(ll j=i,cn=1;j<=m;j+=i,cn++) {
      g[i]=(g[i]+sgm0[j])%p;
    }
  }
  ff[1]=1;
  for(ll i=2;i<=n;i++) {
    if(!ff[i]) {prime[++cnt]=i;}
    for(ll j=1;j<=cnt&&(long long)i*(long long)prime[j]<=(long long)n;j++) {
      ff[i*prime[j]]=1;if(i%prime[j]==0) break;
    }
  }
  for(ll i=1;i<=cnt;i++) {
    for(ll j=n/prime[i];j;j--) {
      sgm0[j*prime[i]]=(sgm0[j*prime[i]]-sgm0[j]+p)%p;
    }
  }
}

int main() {

  n=read();m=read();p=read();if(n<m) swap(n,m);

  Init();

  for(ll i=1;i<=m;i++) {
    ll tmp=((long long)sgm0[i]*(long long)f[i]%(long long)p)
    *(long long)g[i]%(long long)p;
    ans=(ans+tmp)%p;
  }

  write(ans);

  return 0;
}