题目链接：uva 11290 - Gangs

题目大意：给出n和k，表示要构造一个长度为2*n-2的字符串，OG序列为k的字符串（类似于出栈入栈）。

• 如果字符s2先回到原点（即栈空），那么s2 OG s1
• 如果s1和s2同事回答原点，那么忽略头尾的ES进行比较
• 如果s1和s2的前t个相同，扣掉前t个字符考虑

解题思路：出栈入栈的个数是卡特兰数，每次考虑两个部分
Sstr1Estr2.

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 20;

int f[maxn], sf[maxn][maxn];

void init () {
    f[0] = 1;
    for (int i = 1; i < maxn; i++) {
        sf[i][0] = f[i-1] * f[0];
        for (int j = 1; j < i; j++)
            sf[i][j] = sf[i][j-1] + f[i-j-1] * f[j];
        f[i] = sf[i][i-1];
    }

    /*
    for (int i = 0; i < maxn; i++)
        printf("%d\n", f[i]);
        */
}

char* solve (char* s, int n, int m) {
    if (n == 0)
        return s;

    if (n == 1) {
        *s++ = ‘E‘;
        *s++ = ‘S‘;
        return s;
    }

    int k = upper_bound(sf[n], sf[n]+n, m) - sf[n];

    if (k)
        m -= sf[n][k-1];

    int q = m / f[k], r = m % f[k];

    *s++ = ‘E‘;
    s = solve(s, n - k - 1, q);
    *s++ = ‘S‘;
    s = solve(s, k, r);
    return s;
}

int main () {
    init();
    int n, m;
    while (scanf("%d%d", &n, &m) == 2 && n + m) {
        n--;
        m--;
        if (m >=  f[n])
            printf("ERROR\n");
        else {
            char s[maxn*2];
            *solve(s, n, m) = ‘\0‘;
            printf("%s\n", s);
        }
    }
    return 0;
}