如果知道STL函数库中的next_permutation函数，此题丝毫没有难度。然后可以反复调用此函数即可。我在此基础上做了优化。因为第N!th的序列都相当于将序列的后n位逆序排列。所以可以在调用next_permutation函数之前，优化前N!次调用(N!<nth)。

#include 
    
     #include 
     
      using namespace std;#include
      
       int main(){    //freopen("Ignatius and the Princess II.txt","r",stdin);    int seqLength,seqNumber;    while(scanf("%d%d",&seqLength,&seqNumber)!=EOF)    {        vector
       
         array;        int sum=1,start=1;        while(seqNumber>=sum*(start+1))        {            start++;            sum*=start;        }        for (int i=1;i<=seqLength;i++)            array.push_back(i);        int reverseStart=seqLength-start;        reverse(array.begin()+reverseStart,array.end());        for (int i=sum+1;i<=seqNumber;i++)            next_permutation(array.begin(),array.end());        for (int i=0;i