我们设最后答案为 x , 我们我们就能用x表示出所有节点下面的苹果个数, 然后用叶子节点求lcm, 取最大的可行解。
#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using namespace std; const int N = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-; int n, a[N];
LL dv[N], sum[N], all;
vector<int> leaf;
vector<int> G[N]; void solve(int u, int fa) {
if(u != && SZ(G[u]) == ) leaf.push_back(u);
sum[u] = a[u];
for(int v : G[u]) {
if(v == fa) continue;
solve(v, u);
sum[u] += sum[v];
}
} void dfs(int u, int fa) {
if(dv[u] > all) {
printf("%lld\n", all);
exit();
}
for(int v : G[u]) {
if(v == fa) continue;
if(u == ) dv[v] = dv[u] * SZ(G[u]);
else dv[v] = dv[u] * (SZ(G[u]) - );
dfs(v, u);
}
} bool check(LL x) {
for(auto& id : leaf) {
if(x / dv[id] > a[id]) return false;
}
return true;
}
int main() {
scanf("%d", &n);
for(int i = ; i <= n; i++) scanf("%d", &a[i]), all += a[i];
for(int i = ; i < n; i++) {
int u, v; scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
dv[] = ;
solve(, );
dfs(, );
LL lcm = ;
for(auto& id : leaf) {
LL gcd = __gcd(lcm, dv[id]);
if(1.0 * lcm / gcd * dv[id] > all) {
printf("%lld\n", all);
return ;
}
lcm = lcm / gcd * dv[id];
}
LL low = , high = all / lcm, ans = ;
while(low <= high) {
LL mid = low + high >> ;
if(check(mid * lcm)) ans = mid, low = mid + ;
else high = mid - ;
}
printf("%lld\n", all - ans * lcm);
return ;
} /*
*/