YbtOJ 栈的问题

很显然这是一道递推题，两种思路来做

1.如果能一眼看出来这是卡特兰数的话，直接用卡特兰数做就行了

#include <bits/stdc++.h>
using namespace std;
inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + ch - 48;
        ch = getchar();
    }
    return x * f;
}
long long n;
long long f[20];
int main() {
    n = read();
    f[0] = f[1] = 1;
    f[2] = 2;
    for (register int i(3); i <= n; i = -~i) {
        for (register int j(0); j < i; j = -~j) {
            f[i] += f[j] * f[i - 1 - j];
        }
    }
    printf("%lld", f[n]);
    return 0;
}

2.根据题目给的条件进行递推

设f[i][j]表示有i个数未输出，j个数未入栈，显然，当j=0的时候，只能按顺序出栈，f[i][j] = 1；

当j≠0时，若选择弹出栈中的数，有f[i-1][j]种可能，若选择将一个数压进栈，有f[i][j-1]种可能

于是递推式：f[i][j] = f[i-1][j] + f[i][j-1]

#include <bits/stdc++.h>
using namespace std;
inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + ch - 48;
        ch = getchar();
    }
    return x * f;
}
long long n;
long long f[20][20];
int main() {
    n = read();
    for (register int i(1); i <= n; i = -~i) f[i][0] = 1;
    for (register int i(1); i <= n; i = -~i) {
        for (register int j(1); j <= i; j = -~j) {
            f[i][j] = f[i - 1][j] + f[i][j - 1];
        }
    }
    printf("%lld", f[n][n]);
    return 0;
}