cf1334 E. Divisor Paths

Link

Link

Solution

每次在图上走动一条边，要么乘以一个素数，要么除以一个素数

如果$u|v$的话，显然我按照每次乘以一个素数的走法，走出来的边权和就是$\frac{v}{u}$的约数个数再减去$1$，而且这个肯定是最短的；如果$u\nmid v$而且$v\nmid u$，那我可以先让$u$走到$gcd(u,v)$，然后再从$gcd(u,v)$走到$v$

从$a$走到其约数$b$的最短路条数和从$\frac{a}{b}$走到$1$的最短路条数是相同的，都是$\frac{a}{b}$的素约数的排列数，而且这是一个带有重复元素的排列组合。

Code

#include 
#include 
#include 
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define maxn 100010
#define eps 1e-8
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
#define mod 998244353ll
struct EasyMath
{
    ll prime[maxn], phi[maxn], mu[maxn];
    bool mark[maxn];
    ll fastpow(ll a, ll b, ll c)
    {
        ll t(a%c), ans(1ll);
        for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
        return ans;
    }
    void exgcd(ll a, ll b, ll &x, ll &y)
    {
        if(!b){x=1,y=0;return;}
        ll xx, yy;
        exgcd(b,a%b,xx,yy);
        x=yy, y=xx-a/b*yy;
    }
    ll inv(ll x, ll p)  //p是素数
    {return fastpow(x%p,p-2,p);}
    ll inv2(ll a, ll p)
    {
        ll x, y;
        exgcd(a,p,x,y);
        return (x+p)%p;
    }
    void shai(ll N)
    {
        ll i, j;
        for(i=2;i<=N;i++)mark[i]=false;
        *prime=0;
        phi[1]=mu[1]=1;
        for(i=2;i<=N;i++)
        {
            if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
            for(j=1;j<=*prime and i*prime[j]<=N;j++)
            {
                mark[i*prime[j]]=true;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    ll CRT(vector<ll> a, vector<ll> m) //要求模数两两互质
    {
        ll M=1, ans=0, n=a.size(), i;
        for(i=0;i<n;i++)M*=m[i];
        for(i=0;i<n;i++)(ans+=a[i]*(M/m[i])%M*inv2(M/m[i],m[i]))%=M;
        return ans;
    }
}em;
vector<ll> p, fac, plis[maxn];
map<ll,ll> tb;
ll fact[maxn], _fact[maxn], ans[maxn];
int main()
{
    ll D = read(), i, u, v;
    fact[0]=_fact[0]=1;
    rep(i,1,maxn-1)fact[i]=fact[i-1]*i%mod, _fact[i]=em.inv(fact[i],mod);
    for(i=1;i*i<=D;i++)
    {
        if(D%i==0)
        {
            fac.emb(i);
            if(D/i!=i)fac.emb(D/i);
        }
    }
    ll d = D;
    for(i=2;i*i<=d;i++)
    {
        if(d%i==0)
        {
            p.emb(i);
            while(d%i==0)d/=i;
        }
    }
    if(d>1)p.emb(d);
    sort(p.begin(),p.end());
    sort(fac.begin(),fac.end());
    rep(i,0,fac.size()-1)tb[ fac[i] ] = i;
    for(auto x:fac)
    {
        for(auto pr:p)
        {
            if((D/x)%pr==0)
            {
                ll to = tb[x*pr];
                if(plis[to].empty())
                {
                    for(auto d:plis[tb[x]])plis[to].emb(d);
                    plis[to].emb(pr);
                }
            }
        }
    }
    rep(i,0,fac.size())sort(plis[i].begin(),plis[i].end());
    rep(i,0,fac.size()-1)
    {
        auto& vec = plis[i];
        auto& Ans = ans[i];
        Ans = fact[vec.size()];
        ll last = -1, cnt=0;
        for(auto x:vec)
        {
            if(x==last)cnt++;
            else
            {
                Ans = (Ans * _fact[cnt]) %mod;
                cnt = 1;
            }
            last = x;
        }
        Ans = (Ans * _fact[cnt]) %mod;
    }
    ll Q = read();
    while(Q--)
    {
        ll u = read(), v = read(), g = __gcd(u,v);
        printf("%lld\n",ans[tb[u/g]]*ans[tb[v/g]]%mod);
    }
    return 0;
}