HDU 1027 Ignatius and the Princess II

原题代号：HDU 1027

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1027

原题描述：

Ignatius and the Princess II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8415    Accepted Submission(s): 4955

Problem Description

Now our hero finds the door to the BEelzebub feng5166. He opens the door and finds feng5166 is about to kill our pretty Princess. But now the BEelzebub has to beat our hero first. feng5166 says, "I have three question for you, if you can work them out, I will release the Princess, or you will be my dinner, too." Ignatius says confidently, "OK, at last, I will save the Princess."

"Now I will show you the first problem." feng5166 says, "Given a sequence of number 1 to N, we define that 1,2,3...N-1,N is the smallest sequence among all the sequence which can be composed with number 1 to N(each number can be and should be use only once in this problem). So it's easy to see the second smallest sequence is 1,2,3...N,N-1. Now I will give you two numbers, N and M. You should tell me the Mth smallest sequence which is composed with number 1 to N. It's easy, isn't is? Hahahahaha......"
Can you help Ignatius to solve this problem?

 

Input

The input contains several test cases. Each test case consists of two numbers, N and M(1<=N<=1000, 1<=M<=10000). You may assume that there is always a sequence satisfied the BEelzebub's demand. The input is terminated by the end of file.

 

Output

For each test case, you only have to output the sequence satisfied the BEelzebub's demand. When output a sequence, you should print a space between two numbers, but do not output any spaces after the last number.

 

Sample Input

6 4

11 8

 

Sample Output

1 2 3 5 6 4

1 2 3 4 5 6 7 9 8 11 10

 

题目大意：求出1-N个数排列当中第M小的排列

没什么好说的，自己找的个规律，然后模拟了一下操作，所以就直接贴代码了

AC代码：

# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL;

int ans[10]={1},num[10],sum,a[1005];

void swap_(int &b,int &c)
{
    int t;
    t=b,b=c,c=t;
}

int main()
{
    for(int i=1;i<=8;i++)
    {
        ans[i]=ans[i-1]*i;
    }
/*    for(int i=1;i<=10;i++)
        cout<<ans[i]<<endl;*/
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
            a[i]=i;
        mem(num,0);
        sum=0;
        while(m>1)
        {
            for(int i=8;i>=1;i--)
                if(m>ans[i])
                {
                    m-=ans[i],num[i]++,sum++;
                    break;
                }
        }
/*        cout<<sum<<endl;
        for(int i=1;i<=8;i++)
            printf("%2d,,%d\n",i,num[i]);*/
        while(sum)
        {
            for(int i=8;i>=1;i--)
            {
                if(num[i])
                {
                    swap_(a[n-i],a[n-i+num[i]]);
                    sort(a+n-i+1,a+n+1);
                    sum-=num[i];
                    num[i]=0;
                }
            }
        }
        for(int i=1;i<=n;i++)
            printf(i==1?"%d":" %d",a[i]);
        cout<<endl;
    }
    return 0;
}