main
黄海 2 years ago
parent 6a8653ce10
commit aa4f40c316

@ -14,28 +14,31 @@ int n, mod;
vector<int> val[N], pre[N], suc[N];
void dfs1(int u, int fa) {
int tmp = 1;
val[u].push_back(1);
f[u] = 1; // 底为1方便累乘
val[u].push_back(1); // 用一个动态数组,记录都是哪些数字在累乘。
// f[u]是累乘并取模的结果val[u]是都是哪些数字累乘的,一个是结果,一个是过程
for (int i = h[u]; ~i; i = ne[i]) {
int v = e[i];
if (v == fa) continue;
dfs1(v, u);
son[u]++;
val[u].push_back(f[v] + 1);
tmp = tmp * (f[v] + 1) % mod;
son[u]++; // 记录u有多少个儿子,注意不是习惯的有多少个子孙节点,而是儿子有多少个
val[u].push_back(f[v] + 1); // 记录过程
f[u] = f[u] * (f[v] + 1) % mod; // 记录结果
}
f[u] = tmp;
}
void dfs2(int u, int fa) {
if (u == 1) g[u] = 1;
ans[u] = f[u] * g[u] % mod;
pre[u].resize(son[u] + 2);
suc[u].resize(son[u] + 2);
pre[u][0] = suc[u][son[u] + 1] = 1;
for (int i = 1; i <= son[u]; i++) pre[u][i] = (ll)pre[u][i - 1] * val[u][i] % mod;
for (int i = son[u]; i >= 1; i--) suc[u][i] = (ll)suc[u][i + 1] * val[u][i] % mod;
// 前缀积
for (int i = 1; i <= son[u]; i++) pre[u][i] = pre[u][i - 1] * val[u][i] % mod;
// 后缀积
for (int i = son[u]; i >= 1; i--) suc[u][i] = suc[u][i + 1] * val[u][i] % mod;
int m = 0;
for (int i = h[u]; ~i; i = ne[i]) {
int v = e[i];
if (v == fa) continue;
@ -56,6 +59,7 @@ signed main() {
add(a, b), add(b, a);
}
dfs1(1, 0);
g[1] = 1;
dfs2(1, 0);
for (int i = 1; i <= n; i++) cout << ans[i] << endl;
}
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