n the number of sequentially into the stack, the number of stack order?

For each number, you must go to the stack once and out of the stack. We set the stack to state ' 1 ' and the stack is set to state ' 0 '. All States of the n number correspond to a 2n binary number consisting of n 1 and N 0.　　Because the operands waiting for the stack are arranged in the order of 1‥n, the number of operands in the stack is greater than or equal to the operand a (a≤b) of the stack, so the total number of output sequences = the number of binary numbers of 2n bits consisting of n 1 and N 0, and 1 of the cumulative number of the cumulative number not less than 0. In the 2n bit binary number is filled with N 1 scheme number is C (2n,n), do not fill 1 of the remaining n bits are automatically filled 0. The number of scenarios from which non-conforming requirements are subtracted (from left to right scanning, 0 cumulative number greater than 1) is sought. The characteristics of the non-conforming number are scanned by left and right, the cumulative number of m+1 0 and the cumulative number of M 1 must first appear on an odd digit 2m+1 bit, and thereafter 2 (N-M) 1 bits have n-m 1 and n-m-1 0.　　If the back of this 2 (n-m)-1 bits on the 0 and 1 interchange, so that it becomes n-m 0 and N-m-1 1, the result is 1 by the n+1 0 and N-1 1 is composed of 2n digits, that is, an undesirable number corresponds to a n+1 0 and N-1 1 is composed of an arrangement. In turn, any one consists of n+1 0 and n-1 1 of the binary number of 2n, because 0 of the number of 2, 2n is even, it must be on an odd bit of a 0 cumulative number of more than 1 cumulative number. Also in the following sections 0 and 1 are interchangeable, so that it is made up of n 0 and N 1 composed of 2n digits, that is, n+1 0 and n-1 1 composed of 2n digits must correspond to a non-conforming number. Thus the non-required 2n digits correspond to the permutation one by one composed of n+1 0,n-1 1. Clearly, the number of non-conforming schemes is C (2n,n+1). This results in the total number of output sequences =c (2n,n)-C (2n,n+1) =c (2n,n)/(n+1) =h (n+1).

The simulation process is as follows, DFS to populate the stack and the flag out of the stack
#include <iostream>#include<cstring>#include<cstdio>#include<stack>#defineN 100using namespacestd;intA[n];intp[2*N];intpre[2*N];intused[2*N];intindex[2*n];//if a[i]==1, then the number of index[i] is in the stackintN;intCnt//record how many out-of-stack sequencesintCintN) {    intans=1; intm =2*N; intnn =N;  for(intI=1; i<=n; ++i) {ans*= m--; Ans/=i; }    returnans/(nn+1);}voidsolve () {memset (PRE,0,sizeof(pre)); memset (Used,0,sizeof(used)); intPreK =0;  for(intI=1; i<=2*n; ++i) {        if(A[i] = =1) {Pre[i]=PreK; PreK=i; } Else {            intII =PreK;  while(Used[ii]) II=Pre[ii]; cout<<p[index[ii]]<<" ";//out of the stackUSED[II] =1; Pre[prek]=Pre[ii]; }} cout<<Endl;}voidsolve1 () {stack<int>s; intk=1;  for(intI=1; i<=2*n; ++i) {        if(A[i] = =1) S.push (p[k++]); Else{cout<<s.top () <<" ";        S.pop (); }} cout<<Endl;}voidDfsintKintCnt0,intcnt1) {    if(k>2*N) {        //solve ();solve1 (); ++CNT; return ; }    if(cnt1<n) {//into the stackA[K] =1; INDEX[K]= cnt1+1;//number of steps into the stackDFS (k +1, Cnt0, cnt1+1); }    if(CNT0&LT;CNT1) {//out of the stackA[K] =0; DFS (k+1, cnt0+1, Cnt1); }}intMain () {CIN>>N;  for(intI=1; i<=n; ++i) Cin>>P[i]; DFS (1,0,0); cout<<endl<<"number of simulations:"<<cnt<<Endl; cout<<"number of formulas:"<<c (n) <<Endl; return 0;}

Similar questions buy changeA 2n person lined up to enter the theater. Admission fee is 5 yuan. of which only n individuals have a 5-dollar bill, the other N people only 10 yuan banknotes, the theater no other banknotes, asked how many of the methods so long as there are 10 of people buy tickets, the ticket office has 5 yuan of money change? (will hold 5 yuan to reach the view as 5 yuan into the stack, holding 10 yuan to see as a stack of some 5 yuan out of the stack) The final result: C (2n,n)-C (2n,n+1)

n the number of sequentially into the stack, the number of stack order?