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HDU 1730 Northcott Game
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[题目传送门](https://acm.hdu.edu.cn/showproblem.php?pid=1730)
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图1<center><img src='https://acm.hdu.edu.cn/data/images/C63-1003-1.png'></center>
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图2
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<center><img src='https://acm.hdu.edu.cn/data/images/C63-1003-2.png'></center>
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### 一、解题思路
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* 把同一行棋子之间的距离看做石子数。两个棋子紧挨着，就表示这堆石子个数为零。否则石子数量就是白色棋子坐标与黑色棋子坐标差+1。
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* 如果黑棋选择扩大距离(向左走)
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白棋足够聪明，直接跟进，贴上黑棋，这样，本行黑棋不管怎么操作，都会被跟进，直到遇到左侧边界，那么，黑棋将在本行无路可走,只能再去其它行尝试，也就是黑棋在本行没有占到便宜，被迫进入下一行。换句话说，在本题中不能扩大距离，只能缩小距离，即拿走一些石子。
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* 如果黑棋选择缩小距离(向右走)
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对比Nim游戏，就是拿走一些石子，一共有N堆石子，每次只能从某一堆中拿走一些石子（`>=1 && <=a[i]`)个。问是先手必胜，还是先手必败，这不就是经典的Nim游戏吗？
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直接以坐标差$+1$做为石子个数，构建$Nim$游戏，计算异或和，是$0$则先手必败，否则先手必胜。
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### 二、模拟第二组数据， 黑棋是怎么赢的
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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int main() {
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int res = 2 ^ 0 ^ 0;
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printf("%d\n", res); //异或和
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printf("%d\n", res ^ 2); //剩余几个 输出0
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printf("%d\n", 2 - (res ^ 2)); //取走几个 输出2
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return 0;
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}
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```
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### 三、实现代码
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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int main() {
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int n, m;
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while (~scanf("%d %d", &n, &m)) {
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int res = 0;
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for (int i = 1; i <= n; ++i) {
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int a, b;
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scanf("%d %d", &a, &b);
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res = res ^ (abs(a - b) - 1);
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}
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if (res == 0)
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puts("BAD LUCK!");
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else
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puts("I WIN!");
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}
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return 0;
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}
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```
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