From 1155284ae88ce39b70d94917f0ee5acd1228cadb Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Tue, 26 Dec 2023 11:03:30 +0800
Subject: [PATCH] 'commit'

---
 TangDou/AcWing_TiGao/T5/GameTheory/1321.cpp | 45 ++++++++++++---------
 TangDou/AcWing_TiGao/T5/GameTheory/1321.md  |  6 +--
 2 files changed, 26 insertions(+), 25 deletions(-)

diff --git a/TangDou/AcWing_TiGao/T5/GameTheory/1321.cpp b/TangDou/AcWing_TiGao/T5/GameTheory/1321.cpp
index 67a2816..7f6502d 100644
--- a/TangDou/AcWing_TiGao/T5/GameTheory/1321.cpp
+++ b/TangDou/AcWing_TiGao/T5/GameTheory/1321.cpp
@@ -1,38 +1,43 @@
-#include <iostream>
-#include <cstring>
+#include <bits/stdc++.h>
 using namespace std;
 const int N = 55, M = 50060;
 int n, f[N][M];
-int sg(int a, int b) {
-    if (f[a][b] != -1) return f[a][b];
-    int &s = f[a][b];
-    if (!a) return b & 1; // 如果能转移到必败态
-    if (b == 1) return sg(a + 1, b - 1);
-    if (a && !sg(a - 1, b)) return s = 1;                          // 取a
-    if (b && !sg(a, b - 1)) return s = 1;                          // 合并b,取b
-    if (a && b > 1 && !sg(a - 1, b + 1)) return s = 1;             // 合并a,b
-    if (a > 1 && !sg(a - 2, b == 0 ? b + 2 : b + 3)) return s = 1; // 合并a
-    return s = 0;
+
+int dfs(int a, int b) {
+    if (~f[a][b]) return f[a][b]; // 记忆化搜索
+    int &v = f[a][b];             // 引用，赋值更快捷
+    if (!a) return b & 1;         // a==0时，看b是不是奇数，奇数必胜，否则必败
+
+    if (b == 1) return dfs(a + 1, b - 1);
+    if (a && !dfs(a - 1, b)) return v = 1;                          // 从左边取一石子:左侧不空，并且取完后整体是一个必败态,那我必胜
+    if (b && !dfs(a, b - 1)) return v = 1;                          // 合并b
+    if (a && b > 1 && !dfs(a - 1, b + 1)) return v = 1;             // 合并a,b各一个
+    if (a > 1 && !dfs(a - 2, b == 0 ? b + 2 : b + 3)) return v = 1; // 合并a
+    return v = 0;
 }
 int main() {
     int T;
-    scanf("%d", &T);
-    memset(f, -1, sizeof f);
+    cin >> T;
+    memset(f, -1, sizeof f); // 初始化DP数组，-1
+
+    // 边界初始化
     f[1][0] = f[2][0] = 1;
     f[3][0] = 0;
+
     while (T--) {
-        scanf("%d", &n);
+        cin >> n;
         int a = 0, b = -1;
         for (int i = 1; i <= n; i++) {
             int x;
-            scanf("%d", &x);
+            cin >> x;
             if (x == 1)
-                a++;
+                a++; // 左侧石子个数为1的石子堆数量
             else
-                b += x + 1;
+                b += x + 1; // 1:新增加1堆，x:这一堆x个
         }
-        if (b < 0) b = 0;
-        if (sg(a, b))
+        if (b < 0) b = 0; // 一堆都没有，b=0
+
+        if (dfs(a, b))
             puts("YES");
         else
             puts("NO");
diff --git a/TangDou/AcWing_TiGao/T5/GameTheory/1321.md b/TangDou/AcWing_TiGao/T5/GameTheory/1321.md
index e47be1e..e6d5553 100644
--- a/TangDou/AcWing_TiGao/T5/GameTheory/1321.md
+++ b/TangDou/AcWing_TiGao/T5/GameTheory/1321.md
@@ -105,12 +105,8 @@ NO
 
 <center><img src='https://cdn.acwing.com/media/article/image/2021/05/15/61813_30ead721b5-image-20210515211400052.png'></center>
 
-$Q:$**情况**$3$**为什么是两个表达式？**
-答： 
-①当右侧存在时，合并左边两堆石子，则右侧多出一堆石子，并且，石子个数增加$2$,也就是$b+=3$
-
-②当右侧一个都没有的时候，左边送来了一堆，两个石子，按$b$的定义，是堆数+石子个数$-1=2$，即$b+=2$
 
+![](https://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/202312261053532.png)
 
 
 ### 六、实现代码