标签：will feng5166 sequence int number next Hdu1027 permutation Princess
Ignatius and the Princess II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14475 Accepted Submission(s): 8296
"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 理解题意： 第一个数是从1-N一共有这么几个数，然后是把这几个数全排列，输出第二个数打的排序 解题思路： 用next_permutation()实现全排列，在输出 代码：
#include <bits/stdc++.h>
using namespace std;
int a[1001];
int main()
{
int n,m;
while(cin>>n>>m){
for(int i = 1;i <= n; i++)
a[i] = i;
int b = 1;
do{
if(b==m)
break;
b++;
}while(next_permutation(a+1,a+n+1));
for(int i = 1;i < n; i++)
cout<<a[i]<<" ";
cout<<a[n]<<endl;
}
return 0;
}
标签：will,feng5166,sequence,int,number,next,Hdu1027,permutation,Princess 来源： https://www.cnblogs.com/jingshixin/p/12241799.html
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