其实可以用$sum(i,j)$表示从$i$到$1$的$k$次方的值,然后就是$lca$的基本操作
注意,能一起干的事情就一起搞,要不会超时
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #define int long long #define mod 998244353 using namespace std; const int N=300001; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } struct node{ signed u,v,nex; }x[N<<1]; signed n,m,cnt; int sum[N][51]; signed head[N],deep[N],fa[N][20]; void add(signed u,signed v){ x[cnt].u=u,x[cnt].v=v,x[cnt].nex=head[u],head[u]=cnt++; } int ksm(int a,int b){ if(a==1) return 1; int ans=1; a%=mod; while(b){ if(b&1) ans*=a,ans%=mod; a*=a,a%=mod; b>>=1; }return ans; } void dfs(signed f,signed fath){ deep[f]=deep[fath]+1,fa[f][0]=fath; for(signed i=1;i<=50;i++) sum[f][i]=sum[fath][i]+ksm(deep[f],i); for(signed i=1;(1<<i)<=deep[f];++i) fa[f][i]=fa[fa[f][i-1]][i-1]; for(signed i=head[f];i!=-1;i=x[i].nex){ if(x[i].v==fath) continue; dfs(x[i].v,f); } } int Log2[N]; signed lca(signed u,signed v){ if(deep[u]<deep[v]) swap(u,v); for(signed i=Log2[u];i>=0;--i) if(deep[u]-(1<<i)>=deep[v]) u=fa[u][i]; if(u==v) return u; for(signed i=Log2[v];i>=0;--i){ if(fa[u][i]==fa[v][i]) continue; u=fa[u][i],v=fa[v][i]; }return fa[u][0]; } signed q; bool ff; signed main(){ memset(head,-1,sizeof(head)); n=read(); for(signed i=1;i<n;++i){ int u=read(),v=read(); add(u,v),add(v,u); }deep[0]=-1;Log2[0]=1; for(int i=1;i<=n;++i) Log2[i]=Log2[i>>1]+1; dfs(1,0); q=read(); while(q--){ int u=read(),v=read(),k=read(); int ls=lca(u,v); printf("%d ",(((sum[u][k]+sum[v][k]-2*sum[ls][k]+ksm(deep[ls],k))%mod+mod)%mod)); } return 0; }