关于对对碰游戏的研究

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（封面图片来源于网络侵权删）

关于灵感

小学时不知道同学从哪里带回来的游戏，规则是这样的：

两个玩家首先比出两个1的手势，情况是这样的：

a玩家：左手：1，右手：1

b玩家：左手：1，右手：1

然后决定先后顺序，依次在每个自己的回合做出一下行动：使用自己的左手或右手碰一下对方的左手或右手，然后将碰到的两只手所表示的数值相加，自己碰的那只手就变为该数值。如：

原局面：

a玩家：左手：1，右手：1

b玩家：左手：1，右手：1

a玩家使用左手碰了对方的右手后：

a玩家：左手：1+1=2，右手：1

b玩家：左手：1，右手：1

到b玩家时，b玩家可以用左手碰对方的右手：

a玩家：左手：2，右手：1

b玩家：左手：3，右手：1

谁先将两只手都到达目标数字（一般是10）就赢了

我在一次某讲座上和同学玩这个游戏时突然想到，这个游戏有没有必胜的策略？遂作此程序。（不知道这算不算博弈论了反正先当标题党写上吧）

思路

那当然是！dfs直接打爆力求解！

这部分没什么好说的，不会的可以去我的专栏或B站

于是我们得到以下代码：

#include "bits/stdc++.h" #include <conio.h> #include <windows.h> using namespace std; bool handMarker[20][20][20][20]; int endNumber, aLeftHand, aRightHand, bLeftHand, bRightHand; int aMayWin = 0, bMayWin = 0; void color(int m) { HANDLE consolehend; consolehend = GetStdHandle(STD_OUTPUT_HANDLE); SetConsoleTextAttribute(consolehend, m); } void dfs(int aLeftHand, int aRightHand, int bLeftHand, int bRightHand, int step) { // 成功后输出 if ((aLeftHand == endNumber && aRightHand == endNumber) || (bLeftHand == endNumber && bRightHand == endNumber)) { if ((step - 1) % 2 == 0) { aMayWin++; } else { bMayWin++; } // color(0x02); for (int i = 1; i <= step - 1; i++) { printf(u8"│ "); } printf(u8"├─"); printf(u8"%c:Win!:)\n", (step - 1) % 2 + 'a'); for (int i = 1; i <= step - 1; i++) { printf(u8"│ "); } printf(u8"└─"); printf(u8"Used Step: %d.\n", step); // color(0x0F); return ; } handMarker[aLeftHand][aRightHand][bLeftHand][bRightHand] = 1; // 输出 if (step >= 1) { for (int i = 1; i <= step - 1; i++) { printf(u8"│ "); } printf(u8"├─"); printf(u8"%c:My turn!\n", (step + 1) % 2 + 'a'); for (int i = 1; i <= step - 1; i++) { printf(u8"│ "); } printf(u8"├─"); printf(u8"back:a:%d %d b:%d %d\n", aLeftHand, aRightHand, bLeftHand, bRightHand); } int newHand; if (step % 2 == 0) { if (aLeftHand != endNumber) { if (bLeftHand != endNumber) { newHand = (aLeftHand + bLeftHand) % (endNumber + 1); if (newHand == 0) { newHand = 1; } if (!handMarker[newHand][aRightHand][bLeftHand][bRightHand]) { dfs(newHand, aRightHand, bLeftHand, bRightHand, step + 1); } } if (bRightHand != endNumber) { newHand = (aLeftHand + bRightHand) % (endNumber + 1); if (newHand == 0) { newHand = 1; } if (!handMarker[newHand][aRightHand][bLeftHand][bRightHand]) { dfs(newHand, aRightHand, bLeftHand, bRightHand, step + 1); } } } if (aRightHand != endNumber) { if (bLeftHand != endNumber) { newHand = (aRightHand + bLeftHand) % (endNumber + 1); if (newHand == 0) { newHand = 1; } if (!handMarker[aLeftHand][newHand][bLeftHand][bRightHand]) { dfs(aLeftHand, newHand, bLeftHand, bRightHand, step + 1); } } if (bRightHand != endNumber) { newHand = (aRightHand + bRightHand) % (endNumber + 1); if (newHand == 0) { newHand = 1; } if (!handMarker[aLeftHand][newHand][bLeftHand][bRightHand]) { dfs(aLeftHand, newHand, bLeftHand, bRightHand, step + 1); } } } } if (step % 2 == 1) { if (bLeftHand != endNumber) { if (aLeftHand != endNumber) { newHand = (bLeftHand + aLeftHand) % (endNumber + 1); if (newHand == 0) { newHand = 1; } if (!handMarker[aLeftHand][aRightHand][newHand][bRightHand]) { dfs(aLeftHand, aRightHand, newHand, bRightHand, step + 1); } } if (aRightHand != endNumber) { newHand = (bLeftHand + aRightHand) % (endNumber + 1); if (newHand == 0) { newHand = 1; } if (!handMarker[aLeftHand][aRightHand][newHand][bRightHand]) { dfs(aLeftHand, aRightHand, newHand, bRightHand, step + 1); } } } if (bRightHand != endNumber) { if (aLeftHand != endNumber) { newHand = (bRightHand + aLeftHand) % (endNumber + 1); if (newHand == 0) { newHand = 1; } if (!handMarker[aLeftHand][aRightHand][bLeftHand][newHand]) { dfs(aLeftHand, aRightHand, bLeftHand, newHand, step + 1); } } if (aRightHand != endNumber) { newHand = (bRightHand + aRightHand) % (endNumber + 1); if (newHand == 0) { newHand = 1; } if (!handMarker[aLeftHand][aRightHand][bLeftHand][newHand]) { dfs(aLeftHand, aRightHand, bLeftHand, newHand, step + 1); } } } } } int main() { printf("Change to which number will win?"); scanf("%d", &endNumber); // 输入 printf("Beginning number?"); scanf("%d%d%d%d", &aLeftHand, &aRightHand, &bLeftHand, &bRightHand); freopen("out.txt", "w", stdout); printf(u8"back:a:%d %d b:%d %d\n", aLeftHand, aRightHand, bLeftHand, bRightHand); dfs(aLeftHand, aRightHand, bLeftHand, bRightHand, 0); printf(u8"Total Win: a:%d b:%d\n", aMayWin, bMayWin); printf(u8"The possibility of a's winning: %.2f%%\n", 1.0 * aMayWin / (aMayWin + bMayWin) * 100); printf(u8"The possibility of b's winning: %.2f%%\n", 1.0 * bMayWin / (aMayWin + bMayWin) * 100); return 0; }

（做了一点输出优化）

如果你想要在控制台看的话就把main()里的freopen注释掉再把dfs里的color函数取消注释。

结果

很神奇

似乎是一个区间一个区间来的

首先，如果目标数是2的话，a玩家完胜：

Total Win: a:4 b:0 The possibility of a's winning: 100.00% The possibility of b's winning: 0.00%

目标数在3到7的区间里（含3和7）的话，b玩家的胜率先是一下提高，然后缓降：

3： Total Win: a:8 b:11 The possibility of a's winning: 42.11% The possibility of b's winning: 57.89% 4: Total Win: a:21 b:22 The possibility of a's winning: 48.84% The possibility of b's winning: 51.16% 5: Total Win: a:32 b:32 The possibility of a's winning: 50.00% The possibility of b's winning: 50.00% 6: Total Win: a:48 b:42 The possibility of a's winning: 53.33% The possibility of b's winning: 46.67% 7: Total Win: a:86 b:67 The possibility of a's winning: 56.21% The possibility of b's winning: 43.79%

随后在8到11的区间内又缓升：

8: Total Win: a:107 b:98 The possibility of a's winning: 52.20% The possibility of b's winning: 47.80% 9: Total Win: a:140 b:136 The possibility of a's winning: 50.72% The possibility of b's winning: 49.28% 10: Total Win: a:163 b:169 The possibility of a's winning: 49.10% The possibility of b's winning: 50.90% 11: Total Win: a:197 b:211 The possibility of a's winning: 48.28% The possibility of b's winning: 51.72%

而后又降：

12: Total Win: a:244 b:248 The possibility of a's winning: 49.59% The possibility of b's winning: 50.41%

（后面没算了有人跑出来的话发一下评论区谢谢了）

（我猜会不会到后面双方胜率无限接近于50%）

（所以这个游戏最好去重玩，保证不会出现重复局面）

总结

我真无聊，写这个破玩意写到现在，明天还要月考（啊我靠1点了？？？