1027 Ignatius and the Princess II

Ignatius and the Princess II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8885    Accepted Submission(s): 5222

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

6 411 8

 

Sample Output

1 2 3 5 6 41 2 3 4 5 6 7 9 8 11 10
题意： 给你n个数  你的任务是求出他的第m个全排列 。 不理解第m个  就写写就明白了  ，，，先动最右边的数字。
algorithm中有一个求  全排列的函数。  next_permutation  
函数理解  ：  博客： http://blog.csdn.net/lishuhuakai/article/details/8006937
代码：
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define MAXN 100010using namespace std;int n,m;int s[1010];int main(){    while(scanf("%d%d",&n,&m)==2)    {        for(int i=0; i<n; i++)            s[i] = i+1;        int t=1;        bool flag=false;        while(next_permutation(s, s+n))        {            if(flag) break;            t++;            if(t == m)            {                flag = true;                for(int i=0; i<n; i++)                {                    if(i!=0)printf(" ");                    printf("%d",s[i]);                }            }        }        printf("\n");    }}

但是这个题  用个 dfs  也能过。。

代码：
#include <iostream>#include <cstdio>#include <cstring>//注意为什么要用flag标记提前跳出循环。//回溯的原因！！ /**主要原因：1.如果直接return的话就只是跳到上一层，而flag是在回溯之前就定义好的，所以当用flag标记直接调回的时候就不会仅仅返回第上一层，而是直接 逐层返回到第一层，然后结束。 **/ using namespace std ;int n,m,flag;int a[1002],sum,vis[1002];void dfs(int k){    if(flag==1)return;//提前结束     if(k==n+1){        sum++;        if(sum==m){            flag=1;             for(int i=1;i<n;i++){                printf("%d ",a[i]);            }             printf("%d\n",a[n]);            return;//返回上一层         }    }else{        for(int i=1;i<=n;i++){            if(!vis[i]){                vis[i]=1;                a[k]=i;                dfs(k+1);                vis[i]=0;            }        }    }}int main(){    while(~scanf("%d%d",&n,&m)){        memset(vis, 0, sizeof(vis));        sum=flag=0;        dfs(1);        }    return 0;}