Ultraviolet A 11290-Gangs (catlan number)

Ultraviolet A 11290-Gangs (catlan number)

Link to the question: Ultraviolet A 11290-Gangs

Given n and k, it means to construct a string with a length of 2 * N-2, the OG sequence is a string of k (similar to the output stack into the stack ).

• If the character s2 first returns to the origin (that is, the stack is empty), then s2 OG s1
• If s1 and s2 colleagues answer the origin, ignore ES at the beginning and end for comparison.
• If the first t of s1 and s2 are the same, consider

  Solution: the number of outgoing stacks in the stack is catlan. Consider two parts each time.
  Sstr1Estr2.

  #include 
      
       #include 
       
        #include 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;}