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31 lines
1.0 KiB
31 lines
1.0 KiB
## [$AcWing$ $415$. 栈](https://www.acwing.com/problem/content/description/417/)
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(组合计数,卡特兰数) $O(n^2)$
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首先任何一种合法的入栈、出栈操作序列都可以得到一个不同的$1$~$n$的排列，因此可以得到的排列总数等于合法入栈、出栈操作序列的个数。
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将入栈操作记作$0$，出栈操作记作$1$，那么任意一个合法的入栈、出栈操作序列都和$AcWing$ $889$. 满足条件的$01$序列中定义的合法$01$序列一一对应，因此本题的答案和上一题相同，都是第$n$个卡特兰数。
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 40;
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LL c[N][N];
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int main() {
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// 预处理组合数
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for (int i = 0; i < N; i++)
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for (int j = 0; j <= i; j++)
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if (!j)
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c[i][j] = 1;
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else
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c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
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int n;
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cin >> n;
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// 卡特兰数
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cout << c[n * 2][n] / (n + 1) << endl;
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return 0;
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}
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``` |