codeforces716E/715C 树上点分治

https://codeforces.com/contest/715/problem/C

点分治...有很多细节的点分治题

我用了欧拉函数+费马小定理求的非质数乘法逆元

然后发现题目保证和10互质....不需要这么麻烦,其实直接套扩欧求逆元就行了

这题依然是满足可减性的信息,所以每一步容斥一次就行

#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<int,int>
#define all(x) x.begin(),x.end()
#define IO ios::sync_with_stdio(false)
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(a,b) cout<<#a<<'['<<b<<"]="<<a[b]<<endl
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x) for(int i=head[x];i;i=e[i].next)
using namespace std;
const int maxn=1e5+10,maxm=2e5+10;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const double PI=acos(-1.0);
int casn,n,k;
ll val[maxn],pw[maxn],phi,inv[maxn],m;
ll pow_mod(ll a,ll b,ll c=m,ll ans=1) {
	while(b) {
		if(b&1) ans=(a*ans)%c;
		a=(a*a)%c,b>>=1;
	}
	return ans;
}
ll euler(ll n) {
    ll res=n,a=n;
    for(int i=2; i*i<=a; i++) {
        if(a%i==0) {
            res=res/i*(i-1);
            while(a%i==0) a/=i;
        }
    }
    if(a>1) res=res/a*(a-1);
    return res;
}

void init(){
  phi=euler(m);inv[0]=pw[0]=1;
  rep(i,1,n){
    pw[i]=pw[i-1]*10%m;
    inv[i]=pow_mod(pw[i],phi-1);
  }
}
class graph{public:
  struct node{int to,next;ll cost;}e[maxm];
  int head[maxn],nume,n,sz[maxn],maxt,stree[maxn];
  void add(int a,int b,ll c=0){e[++nume]={b,head[a],c};head[a]=nume;}
  int vis[maxn],num[maxn],all,mid;
  void getmid(int now=1,int pre=0){
    sz[now]=1;
    for(int i=head[now];i;i=e[i].next){
      if(e[i].to==pre||vis[e[i].to]) continue;
      getmid(e[i].to,now);
      sz[now]+=sz[e[i].to];
    }
    int tmp=max(sz[now]-1,all-sz[now]);
    if(maxt>tmp) maxt=tmp,mid=now;
  }//base
  ll ans,in[maxn],out[maxn];
  map<ll,int> cnt;
  void init(int n){
    this->n=n,nume=1,mid=0;
    rep(i,1,n) vis[i]=head[i]=0;
  }
  ll getout(int now,int pre,int d){
    ll sum=0;
    for(int i=head[now];i;i=e[i].next){
      int to=e[i].to;
      if(to==pre||vis[to]) continue;
      out[to]=(e[i].cost*pw[d]%m+out[now])%m;
      sum+=getout(to,now,d+1);
    }
    sum+=cnt[out[now]];
    ll tmp=(m-in[now])%m*inv[d]%m;
    if(out[now]==tmp) sum--;
    return sum;
  }
  void getin(int now,int pre,int d){
    ll tmp=(m-in[now])%m*inv[d]%m;
    cnt[tmp]++;
    for(int i=head[now];i;i=e[i].next){
      int to=e[i].to;
      if(vis[to]||to==pre) continue;
      in[to]=(in[now]*10%m+e[i].cost)%m;
      getin(to,now,d+1);
    }
  }
  ll getans(int now,int d=0){
    cnt.clear();
    in[now]=out[now]=d;
    getin(now,0,d!=0);
    return getout(now,0,d!=0);
  }
  void divide(int now){
    vis[now]=1;ans+=getans(now);
    for(int i=head[now];i;i=e[i].next){
      int to=e[i].to;
      if(vis[to]) continue;
      ans-=getans(to,e[i].cost);
      all=sz[to],maxt=n+1;
      getmid(to,now);divide(mid);
    }
  }
  void solve(){
    ans=0;maxt=all=n;
    getmid();divide(mid);
  }
}g;
int main() {IO;
  cin>>n>>m;
  init();
  g.init(n);
  rep(i,2,n){
    int a,b;ll c;cin>>a>>b>>c;a++,b++;
    g.add(a,b,c%m);g.add(b,a,c%m);
  }
  g.solve();
  cout<<g.ans<<endl;
	return 0;
}