【LOJ】#2491. 「BJOI2018」求和

题解

对于50个k都维护一个\(i^k\)前缀和即可

查询的时候就是查询一段连续的区间和，再加上根节点的

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define MAXN 300005
#define pb push_back
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
    }
    while(c >= '0' && c <= '9') {
    	res = res * 10 + c - '0';
    	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) {
    	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 998244353;
int a[55][300005],N,M;
int dep[MAXN],st[MAXN * 2][20],len[MAXN * 2],pos[MAXN],idx;
struct node {
    int to,next;
}E[MAXN * 2];
int head[MAXN],sumE;
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int min_dep(int a,int b) {
    return dep[a] < dep[b] ? a : b;
}
void dfs(int u,int fa) {
    pos[u] = ++idx;
    st[idx][0] = u;
    for(int i = head[u] ; i ; i = E[i].next) {
        int v = E[i].to;
        if(v != fa) {
            dep[v] = dep[u] + 1;
            dfs(v,u);
            st[++idx][0] = u;
        }
    }
}
int lca(int u,int v) {
    u = pos[u];v = pos[v];
    if(u > v) swap(u,v);
    int l = len[v - u + 1];
    return min_dep(st[u][l],st[v - (1 << l) + 1][l]);
}
void Init() {
    read(N);
    int u,v;
    for(int i = 1 ; i < N ; ++i) {
        read(u);read(v);add(u,v);add(v,u);
    }
    for(int i = 1 ; i <= N ; ++i) {
        a[1][i] = i;
        for(int j = 1 ; j <= 50 ; ++j) {
            a[j + 1][i] = mul(a[j][i],i);
            a[j][i] = inc(a[j][i - 1],a[j][i]);
        }
    }
    dfs(1,0);
    for(int j = 1 ; j <= 19 ; ++j) {
        for(int i = 1 ; i <= idx ; ++i) {
            if(i + (1 << j) - 1 > idx) break;
            st[i][j] = min_dep(st[i][j - 1],st[i + (1 << j - 1)][j - 1]);
        }
    }
    for(int i = 2 ; i <= idx ; ++i) len[i] = len[i / 2] + 1;

}
void Solve() {
    int u,v,k,ans;
    read(M);
    for(int i = 1 ; i <= M ; ++i) {
        read(u);read(v);read(k);
        int f = lca(u,v);
        ans = 0;
        ans = inc(ans,inc(a[k][dep[u]],MOD - a[k][dep[f]]));
        ans = inc(ans,inc(a[k][dep[v]],MOD - a[k][dep[f]]));
        if(dep[f]) ans = inc(ans,inc(a[k][dep[f]],MOD - a[k][dep[f] - 1]));
        out(ans);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("7.in","r",stdin);
#endif
    Init();
    Solve();
}