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@ -1,38 +1,43 @@
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#include <iostream>
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#include <cstring>
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 55, M = 50060;
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int n, f[N][M];
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int sg(int a, int b) {
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if (f[a][b] != -1) return f[a][b];
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int &s = f[a][b];
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if (!a) return b & 1; // 如果能转移到必败态
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if (b == 1) return sg(a + 1, b - 1);
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if (a && !sg(a - 1, b)) return s = 1; // 取a
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if (b && !sg(a, b - 1)) return s = 1; // 合并b,取b
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if (a && b > 1 && !sg(a - 1, b + 1)) return s = 1; // 合并a,b
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if (a > 1 && !sg(a - 2, b == 0 ? b + 2 : b + 3)) return s = 1; // 合并a
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return s = 0;
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int dfs(int a, int b) {
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if (~f[a][b]) return f[a][b]; // 记忆化搜索
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int &v = f[a][b]; // 引用，赋值更快捷
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if (!a) return b & 1; // a==0时，看b是不是奇数，奇数必胜，否则必败
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if (b == 1) return dfs(a + 1, b - 1);
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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
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if (b && !dfs(a, b - 1)) return v = 1; // 合并b
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if (a && b > 1 && !dfs(a - 1, b + 1)) return v = 1; // 合并a,b各一个
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if (a > 1 && !dfs(a - 2, b == 0 ? b + 2 : b + 3)) return v = 1; // 合并a
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return v = 0;
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}
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int main() {
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int T;
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scanf("%d", &T);
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memset(f, -1, sizeof f);
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cin >> T;
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memset(f, -1, sizeof f); // 初始化DP数组，-1
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// 边界初始化
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f[1][0] = f[2][0] = 1;
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f[3][0] = 0;
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while (T--) {
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scanf("%d", &n);
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cin >> n;
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int a = 0, b = -1;
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for (int i = 1; i <= n; i++) {
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int x;
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scanf("%d", &x);
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cin >> x;
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if (x == 1)
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a++;
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a++; // 左侧石子个数为1的石子堆数量
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else
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b += x + 1;
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b += x + 1; // 1:新增加1堆，x:这一堆x个
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}
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if (b < 0) b = 0;
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if (sg(a, b))
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if (b < 0) b = 0; // 一堆都没有，b=0
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if (dfs(a, b))
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puts("YES");
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else
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puts("NO");
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