「VMware校园挑战赛」小V的和式

Description

给定 $n,m$ ，求

\[\sum\limits_{x_1=1}^{n}\sum\limits_{x_2=1}^{n}\sum\limits_{y_1=1}^{m}\sum\limits_{y_2=1}^{m}\frac{(x_1y_2+x_2y_1)\ mod\ (max\{x_1,x_2\}max\{ y_1,y_2\})}{gcd(max\{x_1,x_2\},max\{ y_1,y_2\})}
\]

数据范围 $1\le n,m\le 10^9$，答案对 $998244353$ 取模。

Solution

约定

$S_k(n)=\sum\limits_{i=1}^{n} i^k$

若 $n>m$ ，则交换 $n,m$ ，保证 $n\le m$。

拆上下指标

先对上下指标做一下处理，得：

\[[1\le x_1\le n][1\le x_2\le n][1\le y_1\le m][1\le y_2\le m]
\]

\[=([x_1<x_2]+[x_1=x_2]+[x_1>x_2])([y_1<y_2]+[y_1=y_2]+[y_1>y_2])
\]

\[=2[x_1<x_2][y_1<y_2]+2[x_1>x_2][y_1<y_2]+2([x_1<x_2][y_1=y_2]+[x_1=x_2][y_1<y_2])+[x_1=x_2][y_1=y_2]
\]

考虑拆开 $[x_1<x_2][y_1<y_2]$ ：

\[[1\le x_1<x_2][1\le y_1<y_2]
\]

\[=([0\le x_1<x_2]-[0=x_1<x_2])([0\le y_1<y_2]-[0=y_1<y_2])
\]

回带到原来式子，得到：

$2[0\le x_1<x_2][0\le y_1<y_2]$

$-2([0=x_1<x_2][0\le y_1<y_2]+[0\le x_1<x_2][0=y_1<y_2])$

$+2[0=x_1<x_2][0=y_1<y_2]$

$+2[x_1>x_2][y_1<y_2]$

$+2([x_1<x_2][y_1=y_2]+[x_1=x_2][y_1<y_2])$

$+[x_1=x_2][y_1=y_2]$

所以分别求出这 $6$ 种情形即可，下面开始推导。

1. $[0\le x_1<x_2][0\le y_1<y_2]$

即求

\[f_1=\sum\limits_{x_1=0}^{n}\sum\limits_{x_2=x_1+1}^{n} \sum\limits_{y_1=0}^{m} \sum\limits_{y_2=y_1+1}^{m} \frac{(x_1y_2+x_2y_1)\ mod\ (x_2y_2)}{gcd(x_2,y_2)}
\]

\[=\sum\limits_{x_2=1}^{n} \sum\limits_{y_2=1}^{m} \frac{1}{gcd(x_2,y_2)} \sum\limits_{x_1=0}^{x_2-1}\sum\limits_{y_1=0}^{y_2-1} (x_1y_2+x_2y_1)\ mod\ (x_2y_2)
\]

引理1

对于所有的 $a \in [0,y), b\in[0,x)$，$(ax+by)\ mod\ (xy)$ 将集合 $S=\{ (a,b) | 0\le a<y, 0\le b<x\}$ 映射到 $T=\{ k\times gcd(a,b) | 0\le k< \frac{ab}{gcd(a,b)}\}$ ，并且对于 $T$ 中的每一元素恰好有 $gcd(a,b)$ 个原象。

感性理解一下即可，故：

\[f_1=\frac{1}{2}\sum\limits_{x_2=1}^{n} \sum\limits_{y_2=1}^{m} (\frac{x_2^2y_2^2}{gcd(x_2,y_2)}-x_2y_2)
\]

将前半部分拎出来：

\[\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{m} \frac{x^2y^2}{gcd(x,y)}
\]

\[=\sum\limits_{d=1}^{n} d^3 \sum\limits_{i=1}^{n/d}\sum\limits_{j=1}^{m/d}i^2j^2 [gcd(i,j)=1]
\]

\[=\sum\limits_{d=1}^{n}d^3\sum\limits_{k=1}^{n/d} \mu(k)k^4 S_2(\frac{n}{dk})S_2(\frac{m}{dk})
\]

\[=\sum\limits_{T=1}^{n} S_2(\frac{n}{T})S_2(\frac{m}{T})G_1(T)
\]

其中 $G_1(n)=n^3 \sum\limits_{d|n} \mu(d)d$

因此

\[f_1=\frac{1}{2}\sum\limits_{T=1}^{n} S_2(\frac{n}{T})S_2(\frac{m}{T})G_1(T)-\frac{n(n+1)m(m+1)}{8}
\]

2. $[0= x_1<x_2][0\le y_1<y_2]$

即求

\[f_2=\sum\limits_{x_2=1}^{n} \sum\limits_{y_1=0}^{m}\sum\limits_{y_2=y_1+1}^{m} \frac{(x_2y_1)\ mod\ (x_2y_2)}{gcd(x_2,y_2)}
\]

\[=\sum\limits_{x_2=1}^{n} \sum\limits_{y_1=0}^{m}\sum\limits_{y_2=y_1+1}^{m} \frac{x_2y_1}{gcd(x_2,y_2)}
\]

其实还有对称的 $[0\le x_1<x_2][0=y_1<y_2]$ ，推导类似。

3. $[0=x_1<x_2][0=y_1<y_2]$

\[f_3=\sum\limits_{x_2=1}^{n}\sum\limits_{y_2=1}^{m}\frac{0}{gcd(n,m)}=0
\]

4. $[x_1>x_2][y_1<y_2]$

\[f_4=\sum\limits_{x_2=1}^{n}\sum\limits_{x_1=x_2+1}^{n}\sum\limits_{y_1=1}^{m}\sum\limits_{y_2=y_1+1}^{m} \frac{(x_1y_2+x_2y_1)\ mod\ (x_1y_2)}{gcd(x_1,y_2)}
\]

\[=\sum\limits_{x_1=1}^{n}\sum\limits_{y_2=1}^{m}\frac{1}{gcd(x_1,y_2)} \sum\limits_{x_2=1}^{x_1-1}\sum\limits_{y_1=1}^{y_2-1}x_2y_1
\]

\[=\sum\limits_{x_1=1}^{n}\sum\limits_{y_2=1}^{m}\frac{S_1(x_1-1)S_1(y_2-1)}{gcd(x_1,y_2)}
\]

\[=\frac{1}{4}\sum\limits_{d=1}^{n}\frac{1}{d} \sum\limits_{x_1=1}^{n}\sum\limits_{y_2=1}^{m}x_1(x_1-1)y_2(y_2-1)[(x_1,y_2)=d]
\]

\[=\frac{1}{4}\sum\limits_{d=1}^{n} \frac{1}{d}\sum\limits_{x_1=1}^{n/d}\sum\limits_{y_2=1}^{m/d}(x_1^2y_2^2d^3-x_1y_2^2d^2-x_1^2y_2d^2+x_1y_2d)[gcd(x_1,y_2=1)]
\]

\[=\frac{1}{4}\sum\limits_{d=1}^{n}\sum\limits_{t=1}^{n/d}\mu(t)\sum\limits_{x_1=1}^{n/dt}\sum\limits_{y_2=1}^{m/dt}(x_1^2y_2^2d^3t^4-x_1y_2^2d^2t^3-x_1^2y_2d^2t^3+x_1y_1dt^2)
\]

\[=\frac{1}{4}\sum\limits_{T=1}^{n}(S_2(\frac{n}{T})S_2(\frac{m}{T})G_1(T)-(S_1(\frac{n}{T})S_2(\frac{m}{T})+S_2(\frac{n}{T})S_1(\frac{m}{T}))G_2(T)+S_1(\frac{n}{T})S_1(\frac{m}{T})G_3(T))
\]

其中 $G_2(n)=n^2\sum\limits_{d|n}\mu(d)d$

$G_3(n)=n\sum\limits_{d|n}\mu(d)d$

5. $[x_1=x_2][y_1<y_2]$

\[f_5=\sum\limits_{x_2=1}^{n}\sum\limits_{y_1=1}^{m}\sum\limits_{y_2=y_1+1}^{m} \frac{x_2(y_1+y_2)\ mod\ (x_2y_2)}{gcd(x2,y2)}
\]

\[=\sum\limits_{x_2=1}^{n}\sum\limits_{y_1=1}^{m}\sum\limits_{y_2=y_1+1}^{m} \frac{x_2y_1}{gcd(x_2,y_2)}
\]

可以发现 $f_5$ 和 $f_2$ 的式子完全一致，且符号相反，因此可以抵消。

显然对于另一半对称的式子也是如此。

6. $[x_1=x_2][y_1=y_2]$

\[f_6=\sum\limits_{x_1=1}^{n}\sum\limits_{y_1=1}^{m} \frac{0}{gcd(x_2,y_2)}=0
\]

合并

答案为

\[2f_1-2f_2+2f_3+2f_4+2f_5+f_6
\]

\[=2f_1+2f_4
\]

因此我们只需要快速求出 $G_1(n),G_2(n),G_3(n)$ 的前缀和即可。

杜教筛

考虑杜教筛，以 $G_1(n)$ 为例。

$f(n)=n^3\sum\limits_{d|n}\mu(d)d$

记 $S(n)=\sum\limits_{i=1}^{n}f(i)$

则 $S(n)=\sum\limits_{i=1}^{n}i^3\sum\limits_{d|i}\mu(d)d$

$=\sum\limits_{d=1}^{n}\mu(d)d^4S_3(\lfloor \frac{n}{d}\rfloor)$

所以我们需要杜教筛求 $f'(i)=\mu(i)i^4$ 的前缀和。

考虑 $g=ID^4$ ，则 $h(n)=f'*g=[n==1]$

故 $S(n)=1-\sum\limits_{i=2}^{n}i^4 \times S(\lfloor \frac{n}{i}\rfloor)$

实现

时间复杂度：$O(n^{\frac{5}{6}})$

空间复杂度：$O(n^{\frac{2}{3}})$

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize("Ofast")

#pragma GCC optimize("inline")

#pragma GCC optimize("-fgcse")

#pragma GCC optimize("-fgcse-lm")

#pragma GCC optimize("-fipa-sra")

#pragma GCC optimize("-ftree-pre")

#pragma GCC optimize("-ftree-vrp")

#pragma GCC optimize("-fpeephole2")

#pragma GCC optimize("-ffast-math")

#pragma GCC optimize("-fsched-spec")

#pragma GCC optimize("unroll-loops")

#pragma GCC optimize("-falign-jumps")

#pragma GCC optimize("-falign-loops")

#pragma GCC optimize("-falign-labels")

#pragma GCC optimize("-fdevirtualize")

#pragma GCC optimize("-fcaller-saves")

#pragma GCC optimize("-fcrossjumping")

#pragma GCC optimize("-fthread-jumps")

#pragma GCC optimize("-funroll-loops")

#pragma GCC optimize("-fwhole-program")

#pragma GCC optimize("-freorder-blocks")

#pragma GCC optimize("-fschedule-insns")

#pragma GCC optimize("inline-functions")

#pragma GCC optimize("-ftree-tail-merge")

#pragma GCC optimize("-fschedule-insns2")

#pragma GCC optimize("-fstrict-aliasing")

#pragma GCC optimize("-fstrict-overflow")

#pragma GCC optimize("-falign-functions")

#pragma GCC optimize("-fcse-skip-blocks")

#pragma GCC optimize("-fcse-follow-jumps")

#pragma GCC optimize("-fsched-interblock")

#pragma GCC optimize("-fpartial-inlining")

#pragma GCC optimize("no-stack-protector")

#pragma GCC optimize("-freorder-functions")

#pragma GCC optimize("-findirect-inlining")

#pragma GCC optimize("-fhoist-adjacent-loads")

#pragma GCC optimize("-frerun-cse-after-loop")

#pragma GCC optimize("inline-small-functions")

#pragma GCC optimize("-finline-small-functions")

#pragma GCC optimize("-ftree-switch-conversion")

#pragma GCC optimize("-foptimize-sibling-calls")

#pragma GCC optimize("-fexpensive-optimizations")

#pragma GCC optimize("-funsafe-loop-optimizations")

#pragma GCC optimize("inline-functions-called-once")

#pragma GCC optimize("-fdelete-null-pointer-checks")

#include<bits/stdc++.h>

using namespace std;

#define rint register int

#define rep(i,l,r) for(rint i=l;i<=r;i++)

#define per(i,l,r) for(rint i=l;i>=r;i--)

#define ll long long

#define ull unsigned long long

#define pii pair<int,int>

#define pll pair<ll,ll>

#define pb push_back

#define fir first

#define sec second

#define mset(s,t) memset(s,t,sizeof(s))

template<typename T1,typename T2>void ckmin(T1 &a,T2 b){if(a>b)a=b;}

template<typename T1,typename T2>void ckmax(T1 &a,T2 b){if(a<b)a=b;}

template<typename T>T gcd(T a,T b){return b?gcd(b,a%b):a;}

int read(){

  int x=0,f=0;

  char ch=getchar();

  while(!isdigit(ch))f|=ch=='-',ch=getchar();

  while(isdigit(ch))x=10*x+ch-'0',ch=getchar();

  return f?-x:x;

}

const int N=10000005;

const int mod=998244353;

ll qpow(ll a,ll b=mod-2){

  ll res=1;

  a%=mod;

  while(b>0){

    if(b&1)res=res*a%mod;

    a=a*a%mod;

    b>>=1;

  }

  return res;

}

const ll inv2=qpow(2);

const ll inv4=qpow(4);

const ll inv6=qpow(6);

const ll inv8=qpow(8);

const ll inv30=qpow(30);

int mu[N],g1[N],g2[N],g3[N];

bool vis[N];

int pr[N>>3],len,L=1e7;

void init(int n){

  mu[1]=1;

  for(register int i=2;i<=n;i++){

    if(!vis[i])pr[len++]=i,mu[i]=-1;

    for(register int j=0;j<len&&pr[j]*i<=n;j++){

      vis[pr[j]*i]=1;

      if(i%pr[j]==0)break;

      mu[pr[j]*i]=-mu[i];

    }

  }

  rep(i,1,n){

    g1[i]=(g1[i-1]+1ll*i*i%mod*i%mod*i%mod*mu[i])%mod;

    g2[i]=(g2[i-1]+1ll*i*i%mod*i%mod*mu[i])%mod;

    g3[i]=(g3[i-1]+1ll*i*i%mod*mu[i])%mod;

  }

}

ll S1(ll x){

  x%=mod;

  return x*(x+1)%mod*inv2%mod;

}

ll S2(ll x){

  x%=mod;

  return x*(x+1)%mod*(2*x+1)%mod*inv6%mod;

}

ll S3(ll x){

  x%=mod;

  return S1(x)*S1(x)%mod;

}

ll S4(ll x){

  x%=mod;

  return x*(x+1)%mod*(2*x+1)%mod*(3*x*x%mod+3*x-1)%mod*inv30%mod;

}

int n,m;

unordered_map<int,ll>Map1,Map2,Map3;

ll GG1(int n){

  if(n<=L)return g1[n];

  if(Map1[n])return Map1[n];

  ll ans=1;

  for(int i=2,j;i<=n;i=j+1){

    j=n/(n/i);

    ans=(ans-GG1(n/i)*(S4(j)-S4(i-1)+mod))%mod;

  }

  return Map1[n]=(ans%mod+mod)%mod;

}

ll GG2(ll n){

  if(n<=L)return g2[n];

  if(Map2[n])return Map2[n];

  ll ans=1;

  for(int i=2,j;i<=n;i=j+1){

    j=n/(n/i);

    ans=(ans-GG2(n/i)*(S3(j)-S3(i-1)+mod))%mod;

  }

  return Map2[n]=(ans%mod+mod)%mod;

}

ll GG3(ll n){

  if(n<=L)return g3[n];

  if(Map3[n])return Map3[n];

  ll ans=1;

  for(int i=2,j;i<=n;i=j+1){

    j=n/(n/i);

    ans=(ans-GG3(n/i)*(S2(j)-S2(i-1)+mod))%mod;

  }

  return Map3[n]=(ans%mod+mod)%mod;

}

unordered_map<int,ll>map4,map5,map6;

ll G1(int n){

  if(map4[n])return map4[n];

  ll ans=0;

  for(int i=1,j;i<=n;i=j+1){

    j=n/(n/i);

    ans=(ans+(GG1(j)-GG1(i-1)+mod)*S3(n/i))%mod;

  }

  return map4[n]=ans;

}

ll G2(int n){

  if(map5[n])return map5[n];

  ll ans=0;

  for(int i=1,j;i<=n;i=j+1){

    j=n/(n/i);

    ans=(ans+(GG2(j)-GG2(i-1)+mod)*S2(n/i))%mod;

  }

  return map5[n]=ans;

}

ll G3(int n){

  if(map6[n])return map6[n];

  ll ans=0;

  for(int i=1,j;i<=n;i=j+1){

    j=n/(n/i);

    ans=(ans+(GG3(j)-GG3(i-1)+mod)*S1(n/i))%mod;

  }

  return map6[n]=ans;

}

ll f1(){

  ll ans=0;

  for(int i=1,j;i<=n;i=j+1){

    j=min(n/(n/i),m/(m/i));

    ans=(ans+S2(n/i)*S2(m/i)%mod*(G1(j)-G1(i-1)+mod))%mod;

  }

  ans=ans*inv2%mod;

  ans=(ans+mod-S1(n)*S1(m)%mod*inv2%mod)%mod;

  return (ans+mod)%mod;

}

ll f4(){

  ll ans=0;

  for(int i=1,j;i<=n;i=j+1){

    j=min(n/(n/i),m/(m/i));

    ans=(ans+S2(n/i)*S2(m/i)%mod*(G1(j)-G1(i-1)+mod))%mod;

    ans=(ans-(S1(n/i)*S2(m/i)+S2(n/i)*S1(m/i))%mod*(G2(j)-G2(i-1)+mod))%mod;

    ans=(ans+S1(n/i)*S1(m/i)%mod*(G3(j)-G3(i-1)+mod))%mod;

  }

  ans=ans*inv4%mod;

  return (ans+mod)%mod;

}

int main(){

  init(L);

  //n=1e9,m=1e9;

  scanf("%d%d",&n,&m);

  if(n>m)swap(n,m);

  printf("%lld\n",2ll*(f1()+f4())%mod);

  return 0;

}

关于对 $[0= x_1<x_2][0\le y_1<y_2]$ 的彩(推)蛋(导)

qwq 其实是因为我一开始没意料到能相互抵消，所以推了不少，这里就保留一下好了。

\[=\sum\limits_{d=1}^{n} \sum\limits_{x_2=1}^{n/d}\sum\limits_{y_2=1}^{m/d}x_2S_1(y_2d-1)[gcd(x_2,y_2)=1]
\]

\[=\sum\limits_{d=1}^{n}\sum\limits_{k=1}^{n/d}\mu(k)kS_1(\frac{n}{dk})\sum\limits_{y_2=1}^{m/dk}S_1(y_2dk-1)
\]

右边 $y_2$ 的式子，令 $t=m/dk$ ，经化简得

\[\frac{1}{2}(d^2k^2S_2(t)-dkS_1(t))
\]

因此：

\[f_2=\frac{1}{2}\sum\limits_{d=1}^{n}\sum\limits_{k=1}^{n/d}\mu(k)kS_1(\frac{n}{dk})(d^2k^2S_2(\frac{m}{dk})-dkS_1(\frac{m}{dk}))
\]

然后令 $T=dk$ 化简一下就行了。。。