Q:N次正确操作後棧還是為空的操作序列。
#include<bits/stdc++.h>
using namespace std;
int pow1(int n,int x)
{
int m=n;
while(x-->1)
{
m*=n;
}
return m;
}
int isok2(int n,int k )
{
if(((n>>k)&1)==0){return 0;}
int sum=0,i=k+1,m=0;
while(i--)
m=(n>>i)&1;
(m==1)?(sum+=m,1):(sum+=-1,0);
if(sum<0)return 0;
if(sum==0)
i=k+1;
while(i--)
(m==1)?(cout<<"入棧,",1):(cout<<"出棧,",0);
}
cout<<endl;
return 1;
}
void out(int n)
int count=0;
cout<<pow(2,n)<<endl;
for(int i=2;i<pow(2,n);i++)
{
if(isok2(i,n-1))count++;
cout<<count<<endl;
}
void PushPopSequence(int num,int count,int &sum,int *record,int len)
if(count < 0){return ;}//(1)count
if(num == 0)
if(count == 0)
++sum;
for(int i=0;i < len;++i)
if(record[i] == 1) cout<<"入棧"<<" ";
else cout<<"出棧"<<" ";
cout<<endl;
return ;
record[len-num] = 1;
PushPopSequence(num-1,count+1,sum,record,len); //進棧
record[len-num] = 2;
PushPopSequence(num-1,count-1,sum,record,len); //出棧
/*
EG:
111_ X不行就傳回(1)
1101 X不行
1100 Y可以
1011 X
1010 Y
。。。
是以就2種
*/
int PushPopSequence(int num)
int sum = 0;
if(num <= 0 || (num & 1))
return 0;
int count = 0;
int *p = (int *)malloc(sizeof(int)*num);
PushPopSequence(num,count,sum,p,num);
free(p);
return sum;
int main()
//PushPopSequence(10);
int x=0;
cin>>x;
out(x);
算法out版本 就是自然智慧 理論上 32位内 可以計算 但大資料50 100 就挂了
入棧,出棧,入棧,出棧,入棧,出棧,
入棧,出棧,入棧,入棧,出棧,出棧,
入棧,入棧,出棧,出棧,入棧,出棧,
入棧,入棧,出棧,入棧,出棧,出棧,
入棧,入棧,入棧,出棧,出棧,出棧,
5
Hello,C++ world of AnycodeX!