【BJOI 2018】 求和

【题目链接】

           点击打开链接

【算法】

           预处理i^k的前缀和，对于每次询问，树上倍增即可

           时间复杂度 : O(nk + mlog(n))

【代码】

          

#include<bits/stdc++.h>
using namespace std;
#define MAXK 55
#define MAXN 300010
#define MAXLOG 20
const long long P = 998244353;

struct Edge
{
        int to,nxt;
} e[MAXN<<1];
int n,x,i,j,y,q,k,Lca,tot;
int anc[MAXN][MAXLOG],dep[MAXN],head[MAXN],fa[MAXN];
long long p[MAXN][MAXK],s[MAXN][MAXK];
long long ans;

template <typename T> inline void read(T &x)
{
    int f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
template <typename T> inline void write(T x)
{
    if (x < 0)
    {
        putchar('-');
        x = -x;
    }
    if (x > 9) write(x/10);
    putchar(x%10+'0');
}
template <typename T> inline void writeln(T x)
{
    write(x);
    puts("");
}
inline void add(int x,int y)
{
        tot++;
        e[tot] = (Edge){y,head[x]};
        head[x] = tot;
}
inline void dfs(int u)
{
        int i,v;
        anc[u][0] = fa[u];
        for (i = 1; i < MAXLOG; i++)
        {
                if (dep[u] < (1 << i)) break;
                anc[u][i] = anc[anc[u][i-1]][i-1];
        }
        for (i = head[u]; i; i = e[i].nxt)
        {
                v = e[i].to;
                if (fa[u] != v)
                {
                        fa[v] = u;
                        dep[v] = dep[u] + 1;
                        dfs(v);
                }
        }        
}
inline int lca(int x,int y)
{
        int i,t;
        if (dep[x] > dep[y]) swap(x,y);
        t = dep[y] - dep[x];
        for (i = 0; i < MAXLOG; i++)
        {
                if (t & (1 << i))
                        y = anc[y][i];    
        }        
        if (x == y) return x;
        for (i = MAXLOG - 1; i >= 0; i--)
        {
                if (anc[x][i] != anc[y][i])
                {
                        x = anc[x][i];
                        y = anc[y][i];
                }
        }
        return fa[x];
}

int main() 
{
        
        read(n);
        for (i = 1; i < n; i++)
        {
                read(x); read(y);
                add(x,y);
                add(y,x);        
        } 
        for (i = 1; i < n; i++)
        {
                p[i][0] = 1;
                for (j = 1; j < MAXK; j++) p[i][j] = p[i][j-1] * i % P;
        }
        for (i = 1; i < n; i++)
        {
                for (j = 1; j < MAXK; j++) 
                {
                        s[i][j] = (s[i-1][j] + p[i][j]) % P;
                }
        }
        dfs(1);
        read(q);
        while (q--)
        {
                read(x); read(y); read(k);    
                Lca = lca(x,y);
                ans = (s[dep[x]][k] + s[dep[y]][k]) % P;
                ans = ((ans - (2 * s[dep[Lca]][k]) % P) + P) % P;
                ans = (ans + p[dep[Lca]][k] + P) % P;
                writeln(ans);
        }
        
        return 0;
    
}