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题目
给定 n 堆石子，两位玩家轮流操作，每次操作可以取走其中的一堆石子，然后放入两堆规模更小
的石子（新堆规模可以为 0，且两个新堆的石子总数可以大于取走的那堆石子数），最后无法进行操作的人视为失败。
问如果两人都采用最优策略，先手是否必胜。
输入格式
第一行包含整数 n。
第二行包含 n 个整数，其中第 i 个整数表示第 i 堆石子的数量 ai。
输出格式
如果先手方必胜，则输出 Yes。
否则，输出 No。
数据范围
1≤n,ai≤100
输入样例：
2 2 3
输出样例：
Yes
思路
对比于集合nim游戏，其他步骤都一样，只不过此时局面的变化用一个循环来遍历（分成两堆规模更小的石子），并且求其sg值
for(int i = 0; i < x; i ++) for(int j = 0; j <= i; j ++) S.insert(sg(i) ^
sg(j));
代码
#include<iostream> #include<cstring> #include<unordered_set> using namespace
std; const int N = 110; int f[N]; int sg(int x) { if(f[x] != -1) return f[x];
unordered_set<int> S; for(int i = 0; i < x; i ++) for(int j = 0; j <= i; j ++)
S.insert(sg(i) ^ sg(j)); //mex操作 for(int i = 0; ; i ++) if(!S.count(i)) return
f[x] = i; } int main() { int n; cin >> n; memset(f, -1, sizeof f); int res = 0;
for(int i = 0; i < n; i ++) { int x; cin >> x; res ^= sg(x); }
if(res)puts("Yes"); else puts("No"); return 0; }
