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51
TangDou/AcWing_TiGao/T5/GameTheory/1322.cpp
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51
TangDou/AcWing_TiGao/T5/GameTheory/1322.cpp
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#include <cstdio>
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using namespace std;
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const int N = 1010;
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int n;
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int a[N];
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int left[N][N], right[N][N]; // left,right 在 iostream库中用过了,不能用!
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int main() {
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int T;
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scanf("%d", &T);
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while (T--) {
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scanf("%d", &n);
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for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
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for (int len = 1; len <= n; len++) // 枚举长度
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for (int i = 1; i + len - 1 <= n; i++) { // left[i][j],从i到j
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int j = i + len - 1;
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if (len == 1)
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left[i][j] = right[i][j] = a[i]; // DP初始值
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else {
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int L = left[i][j - 1], R = right[i][j - 1], X = a[j];
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if (R == X)
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left[i][j] = 0;
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else if (X < L && X < R || X > L && X > R)
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left[i][j] = X;
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else if (L > R)
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left[i][j] = X - 1;
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else
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left[i][j] = X + 1;
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L = left[i + 1][j], R = right[i + 1][j], X = a[i];
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if (L == X)
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right[i][j] = 0;
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else if (X < L && X < R || X > L && X > R)
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right[i][j] = X;
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else if (R > L)
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right[i][j] = X - 1;
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else
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right[i][j] = X + 1;
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}
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}
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if (n == 1)
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puts("1");
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else
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printf("%d\n", left[2][n] != a[1]);
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}
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return 0;
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}
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