"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?
InputThe 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.
OutputFor 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
题解:求全排列哦
#include<cstdio> #include<iostream> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> typedef long long ll; using namespace std; int n,m,a[1005]; int main(){ while(scanf("%d %d",&n,&m)!=EOF){ for(int i=1;i<1005;i++) a[i]=i; for(int i=1;i<m;i++) next_permutation(a+1, a+1+n); printf("%d",a[1]); for(int i=2;i<=n;i++) printf(" %d",a[i]); printf("\n"); } return 0; }