From aa4f40c316cf1115a2abdbee523fdeb8ee30aef7 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Thu, 18 Jan 2024 10:12:58 +0800
Subject: [PATCH] 'commit'

---
 TangDou/Topic/HuanGenDp/SubTree.cpp | 22 +++++++++++++---------
 1 file changed, 13 insertions(+), 9 deletions(-)

diff --git a/TangDou/Topic/HuanGenDp/SubTree.cpp b/TangDou/Topic/HuanGenDp/SubTree.cpp
index ea3cee7..7332a5c 100644
--- a/TangDou/Topic/HuanGenDp/SubTree.cpp
+++ b/TangDou/Topic/HuanGenDp/SubTree.cpp
@@ -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;
 }
\ No newline at end of file