Enumeration of oj 2030 permutation numbers of Fuzhou University

/* ACMer: The MDK time is the slowest one... Khan. UserID: MDKSubmit time: 17: 29: 16 Language: C ++ Length: 2061 Bytes. result: Accepted */# include <stdio. h> # include <iostream> # include <limits. h> # include <string. h> # include <math. h> # include <vector> # include <algorithm> # include <set> # include <map> # include <stack> # define MAXN 2000 using namespace std; char c [20], a [20]; string ku = "0000000000000000000"; void fun (int n, int r) {if (n) {fun (n/r, r); ku + = n % r> 9? N % R-10 + 'A': n % r + '0';} int next (int j) {for (int I = j + 1; I <strlen (a); I ++) if (a [I]! = '') {// Cout <" next (j): "<I <endl; return I;} return MAXN;} int OK () {for (int I = 0; I <strlen (a); I ++) {for (int j = 0; j <strlen (a); j ++) {if (a [j] = '(' & next (j )! = MAXN & a [next (j)] = ') {a [j] = ''; a [next (j)] = '';}}} for (int I = 0; I <strlen (a); I ++) if (a [I]! = '') Return 0; return 1;} int main () {while (cin> c) {int n = 0, d [20], flag = 0, num = 0; for (int I = 0; I <strlen (c); I ++) {if (c [I] = '? ') D [n] = I, n ++;} if (n = 0) cout <"-0" <endl; else {for (int I = 0; I <pow (2, n); I ++) {fun (flag ++, 2); strcpy (a, c ); for (int j = 0; j <n; j ++) {if (ku [ku. length ()-1-j] = '0') a [d [j] = '('; if (ku [ku. length ()-1-j] = '1') a [d [j] = ')';} if (OK () {num ++ ;} ku = "000000000000000000" ;}cout <num <endl ;}}}

The main idea is enumeration, which is a very stupid method. The Ugly code is stuck.

I found a method to enumerate all the sorting conditions. The main principle is to use the hexadecimal limit.

String ku = "0000000000000000000 ";
Void fun (int n, int r) {// converts every hexadecimal value in the 10 hexadecimal format. Thanks for the simplified code.
If (n ){
Fun (n/r, r );
Ku + = n % r> 9? N % R-10 + 'A': n % r + '0 ';
}
}

Here we use the binary number, because there are only two enumeration cases: (and). There are several hexadecimal values.

Then hash:

1                 fun(flag++,2);
2                 strcpy(a,c);
3                 for(int j = 0;j<n;j++)
4                 {
5                     if(ku[ku.length()-1-j]=='0')
6                         a[d[j]]='(';
7                     if(ku[ku.length()-1-j]=='1')
8                         a[d[j]]=')';
9                 }

The next step is to determine whether the function meets the conditions, which is easy.

Enumeration of permutation numbers.