[BJOI2018]求和

link

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

#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;
}