BJOI 2012 求和

https://www.luogu.org/problemnew/show/P4427

因为这个题，我交了整整两页...

题目不难，预处理+lca

一个奇坑点：预处理sum的时候，你是不能取模的，因为有可能取模之后sum值变小，而实际上不是，由于你后一步还要做减法

因此就会出锅，导致我爆零一整页正确做法是最后取模，虽然我也不知道为什么这样不会爆炸,300000 * 998244353这都死哪儿去了...

真是血一般的教训啊，取模不能乱取

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<climits>
#include<vector>
#include<cstdlib>
#include<ctime>
#include<queue> 
using namespace std;
#define mod 998244353
#define maxn 300000 + 10
typedef long long ll;
inline ll read()
{
    ll ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') op = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        (ans *= 10) += ch - '0';
        ch = getchar();
    }
    return ans * op;
}
ll power(ll a,ll b)
{
    ll ans = 1,res = a;
    while(b)
    {
        if(b & 1)
            (ans *= res) %= mod;
        (res *= res) %= mod;
        b = b >> 1;
    }
    return ans % mod;
}
struct edge
{
    ll to,next;
}e[maxn << 1];
ll fir[maxn],alloc;
void adde(ll u,ll v)
{
    e[++alloc].next = fir[u];
    fir[u] = alloc;
    e[alloc].to = v;
    swap(u,v);
    e[++alloc].next = fir[u];
    fir[u] = alloc;
    e[alloc].to = v;
}
ll sum[maxn][60];
ll dep[maxn],f[maxn][31];
void dfs(ll u,ll fa)
{
    dep[u] = dep[fa] + 1;
    f[u][0] = fa;
    for(int i = 1;i <= 50;i++)
        sum[u][i] = sum[fa][i] + power(dep[u],i);
    for(int i = 1;i <= 20;i++)
        f[u][i] = f[f[u][i - 1]][i - 1];
    for(int i = fir[u];i;i = e[i].next)
    {
        ll v = e[i].to;
        if(v == fa) continue;
        dfs(v,u);
    }
}
ll lca(ll x,ll y)
{
    if(dep[x] < dep[y]) swap(x,y);
    for(int i = 19;i >= 0;i--)
    {
        if(dep[f[x][i]] >= dep[y]) x = f[x][i];
        if(x == y) return x;
    }
    for(int i = 19;i >= 0;i--)
    {
        if(f[x][i] != f[y][i])
        {
            x = f[x][i];
            y = f[y][i];
        }
    }
    return f[x][0];
}
ll n,m; 
int main()
{
    n = read();
    for(int i = 1;i <= n - 1;i++)
    {
        ll u = read(),v = read();
        adde(u,v);
    }
    dep[0] = -1; 
    dfs(1,0); 
    m = read();
    for(int i = 1;i <= m;i++)
    {
        ll u = read(),v = read(),k = read();
        ll to = lca(u,v);
        printf("%lld
" , ((sum[u][k] + sum[v][k]) - (sum[to][k] + sum[f[to][0]][k])) % mod);
    }
}