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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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const int M = 10010;
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int n, k;
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int a[N]; // 一共几种取法,比如一次取2个或5个。
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int f[M]; // SG函数的值
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int res;
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int sg(int x) {
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if (~f[x]) return f[x]; // 记忆化搜索
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unordered_set<int> S;
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for (int i = 0; i < k; i++)
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if (x >= a[i]) S.insert(sg(x - a[i])); // x-s[i]:x的可行路径中终点有哪几个; sg(x-s[i]):这个终点它的sg值
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for (int i = 0;; i++)
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if (!S.count(i)) return f[x] = i;
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}
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int main() {
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memset(f, -1, sizeof f); // 初始化数组值为-1
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cin >> k; // 表示数字集合S中数字的个数
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for (int i = 0; i < k; i++) cin >> a[i];
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cin >> n; // 一共几堆
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// n堆石子,每堆石子都取SG值,然后异或在一起
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for (int i = 0; i < n; i++) {
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int x;
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cin >> x; // 每堆里多少个
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res ^= sg(x);
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}
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if (res)
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puts("Yes"); // 如果不是零,必胜
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else
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puts("No"); // 如果是零,必败
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return 0;
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}
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