P6810 「MCOI-02」Convex Hull 凸包

「MCOI-02」Convex Hull 凸包

题意即求：

\[\sum _ { i = 1 } ^ { n } \sum _ { j = 1 } ^ { m } \sigma _ 0 ( i ) \sigma _ 0 ( j ) \sigma _ 0 ( \gcd ( i , j ) ) \pmod { p } \]

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：

\[\begin{aligned} & \sum _ { i = 1 } ^ { n } \sum _ { j = 1 } ^ { m } \sigma _ 0 ( i ) \sigma _ 0 ( j ) \sigma _ 0 ( \gcd ( i , j ) ) \\ = & \sum _ { d = 1 } \sum _ { i = 1 } ^ { n } \sum _ { j = 1 } ^ { m } \sigma _ 0 ( i ) \sigma _ 0 ( j ) \sigma _ 0 ( d ) \left[ \gcd ( i, j ) = d \right] \\ = & \sum _ { d = 1 } \sigma _ 0 ( d ) \sum _ { i = 1 } ^ { \left\lfloor n / d \right \rfloor } \sum _ { j = 1 } ^ { \left\lfloor m / d \right \rfloor } \sigma _ 0 ( i d ) \sigma _ 0 ( j d ) \left[ \gcd ( i, j ) = 1 \right] \\ = & \sum _ { d = 1 } \sigma _ 0 ( d ) \sum _ { i = 1 } ^ { \left\lfloor n / d \right \rfloor } \sum _ { j = 1 } ^ { \left\lfloor m / d \right \rfloor } \sigma _ 0 ( i d ) \sigma _ 0 ( j d ) \sum _ { k \mid i, k \mid j} \mu ( k ) \\ = & \sum _ { d = 1 } \sigma _ 0 ( d ) \sum _ { k = 1} \mu ( k ) \sum _ { i = 1 } ^ { \left\lfloor n / kd \right \rfloor } \sigma _ 0 ( i kd ) \sum _ { j = 1 } ^ { \left\lfloor m / kd \right \rfloor } \sigma _ 0 ( j kd ) \\ = & \sum _ { T = 1 } \sum _ { d \mid T } \sigma _ 0 ( d ) \mu \left(\dfrac{T}{d}\right) \sum _ { i = 1 } ^ { \left\lfloor n / T \right \rfloor } \sigma _ 0 ( i T ) \sum _ { j = 1 } ^ { \left\lfloor m / T \right \rfloor } \sigma _ 0 ( j T ) \end{aligned} \]

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;
}