[题目链接]
https://codeforces.com/contest/986/problem/E
[算法]
X到Y的路径积 ， 可以转化为X到根的路径积乘Y到根的路径积 ， 除以LCA到根的路径积 ， 再除以LCA父节点到根的路径积
考虑如何计算根到X路径上每个点与Value的GCD之积
不妨对于每个质数P开一个数组cnt[] , 表示根到当前节点P^i有多少个 ， 我们可以在DFS的过程中维护这个数组
将询问离线即可
时间复杂度 ： O(V + NlogN + QlogV^2)
[代码]
#include<bits/stdc++.h>
using namespace std;
#define MAXLOG 30
const int P = 1e9 + 7;
const int MAXN = 1e5 + 10;
const int MAXP = 1e6 + 10;
const int MAXV = 1e7 + 10;
struct edge
{
int to , nxt;
} e[MAXN << 1];
struct info
{
int value , home;
bool type;
} ;
int tot , n , t;
int a[MAXN],depth[MAXN],prime[MAXP],head[MAXN],ans[MAXN],f[MAXV];
int cnt[MAXP][MAXLOG],anc[MAXN][MAXLOG];
vector< info > q[MAXN];
template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
T 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;
}
inline void addedge(int u,int v)
{
t++;
e[t] = (edge){v,head[u]};
head[u] = t;
}
inline int exp_mod(int a,int n)
{
int res = 1 , b = a;
while (n > 0)
{
if (n & 1) res = 1ll * res * b % P;
b = 1ll * b * b % P;
n >>= 1;
}
return res;
}
inline int inv(int x) { return exp_mod(x,P - 2); }
inline void dfs(int u,int fa)
{
depth[u] = depth[fa] + 1;
for (int i = 1; i < MAXLOG; i++)
{
if (depth[u] <= (1 << i)) break;
anc[u][i] = anc[anc[u][i - 1]][i - 1];
}
for (int i = head[u]; i; i = e[i].nxt)
{
int v = e[i].to;
if (v == fa) continue;
anc[v][0] = u;
depth[v] = depth[u] + 1;
dfs(v,u);
}
}
inline int lca(int u,int v)
{
if (depth[u] > depth[v]) swap(u,v);
for (int i = MAXLOG - 1; i >= 0; i--)
{
if (depth[anc[v][i]] >= depth[u])
v = anc[v][i];
}
if (u == v) return u;
for (int i = MAXLOG - 1; i >= 0; i--)
{
if (anc[u][i] != anc[v][i])
u = anc[u][i] , v = anc[v][i];
}
return anc[u][0];
}
inline void modify(int x,int delta)
{
for (int i = 1; 1ll * prime[i] * prime[i] <= x; i++)
{
if (x % prime[i] == 0)
{
int p = 0;
while (x % prime[i] == 0)
{
x /= prime[i];
p++;
}
cnt[i][p] += delta;
}
}
if (x != 1)
{
int pos = lower_bound(prime + 1,prime + tot + 1,x) - prime;
cnt[pos][1] += delta;
}
}
inline int query(int x)
{
int ret = 1;
for (int i = 1; 1ll * prime[i] * prime[i] <= x; i++)
{
if (x % prime[i] == 0)
{
int p = 0;
while (x % prime[i] == 0)
{
p++;
x /= prime[i];
}
int s = 0;
for (int j = 1; j <= p; j++) s += cnt[i][j] * j;
for (int j = p + 1; j < MAXLOG; j++) s += cnt[i][j] * p;
ret = 1ll * ret * exp_mod(prime[i],s) % P;
}
}
if (x != 1)
{
int pos = lower_bound(prime + 1,prime + tot + 1,x) - prime;
int s = 0;
for (int i = 1; i < MAXLOG; i++) s += cnt[pos][i];
ret = 1ll * ret * exp_mod(x,s) % P;
}
return ret;
}
inline void solve(int u,int fa)
{
modify(a[u],1);
for (unsigned i = 0; i < q[u].size(); i++)
{
if (q[u][i].type) ans[q[u][i].home] = 1ll * ans[q[u][i].home] * query(q[u][i].value) % P;
else ans[q[u][i].home] = 1ll * ans[q[u][i].home] * inv(query(q[u][i].value)) % P;
}
for (int i = head[u]; i; i = e[i].nxt)
{
int v = e[i].to;
if (v == fa) continue;
solve(v,u);
}
modify(a[u],-1);
}
int main()
{
read(n);
for (int i = 2; i < MAXV; i++)
{
if (!f[i])
{
f[i] = i;
prime[++tot] = i;
}
for (int j = 1; j <= tot; j++)
{
int tmp = i * prime[j];
if (tmp >= MAXV) break;
f[tmp] = prime[j];
if (prime[j] == f[i]) break;
}
}
for (int i = 1; i < n; i++)
{
int u , v;
read(u); read(v);
addedge(u,v);
addedge(v,u);
}
dfs(1,0);
for (int i = 1; i <= n; i++) read(a[i]);
int Q;
read(Q);
for (int i = 1; i <= Q; i++)
{
int u , v , x;
read(u); read(v); read(x);
int Lca = lca(u,v);
q[u].push_back((info){x,i,true});
q[v].push_back((info){x,i,true});
q[Lca].push_back((info){x,i,false});
if (Lca != 1) q[anc[Lca][0]].push_back((info){x,i,false});
ans[i] = 1;
}
solve(1,0);
for (int i = 1; i <= Q; i++) printf("%d\n",ans[i]);
return 0;
}
[Codeforces 986E] Prince's Problem
原文地址：https://www.cnblogs.com/evenbao/p/9739488.html
时间： 2024-08-01 10:09:28