搜索专题: HDU1027Ignatius 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): 8497 Accepted Submission(s): 5002
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?
file.
6 4
11 8
1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N = 1000 + 5;
bool visit[N];
int a[N],n,m,cnt; void DFS(int cur){
if(cnt>=m) return;
if(cur > n ){
cnt++;
if(cnt == m)
for(int i=1;i<=n;i++)
printf("%d%c",a[i],i<n?' ':'\n');
return;
}
for(int i=1;i<=n;i++){
if(!visit[i]){
visit[i] = true;
a[cur] = i;
DFS(cur + 1);
visit[i] = false;
}
}
}
int main(){
while(scanf("%d %d",&n,&m)==2){
cnt = 0;
DFS(1);
}
}
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