HDOJ 1027 Ignatius and the Princess II

  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 <algorithm>
#include <vector>
using namespace std;
#include<stdio.h>
int main()
{
//freopen("Ignatius and the Princess II.txt","r",stdin);
int seqLength,seqNumber;
while(scanf("%d%d",&seqLength,&seqNumber)!=EOF)
{
vector<int> 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<seqLength-1;i++)
printf("%d ",array[i]);
printf("%d\n",array[seqLength-1]);
}
return 0;
}