Codeforces Round #309 (Div. 2) D 构造

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D. Kyoya and Permutation

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

Let's define the permutation of length n as an array p = [p1, p2, ..., pn] consisting of n distinct integers from range from 1 to n. We say that this permutation maps value 1 into the value p1, value 2 into the value p2 and so on.

Kyota Ootori has just learned about cyclic representation of a permutation. A cycle is a sequence of numbers such that each element of this sequence is being mapped into the next element of this sequence (and the last element of the cycle is being mapped into the first element of the cycle). The cyclic representation is a representation of p as a collection of cycles forming p. For example, permutation p = [4, 1, 6, 2, 5, 3] has a cyclic representation that looks like (142)(36)(5) because 1 is replaced by 4, 4 is replaced by 2, 2 is replaced by 1, 3 and 6 are swapped, and 5 remains in place.

Permutation may have several cyclic representations, so Kyoya defines the standard cyclic representation of a permutation as follows. First, reorder the elements within each cycle so the largest element is first. Then, reorder all of the cycles so they are sorted by their first element. For our example above, the standard cyclic representation of [4, 1, 6, 2, 5, 3] is (421)(5)(63).

Now, Kyoya notices that if we drop the parenthesis in the standard cyclic representation, we get another permutation! For instance, [4, 1, 6, 2, 5, 3] will become [4, 2, 1, 5, 6, 3].

Kyoya notices that some permutations don't change after applying operation described above at all. He wrote all permutations of length n that do not change in a list in lexicographic order. Unfortunately, his friend Tamaki Suoh lost this list. Kyoya wishes to reproduce the list and he needs your help. Given the integers n and k, print the permutation that was k-th on Kyoya's list.

Input
The first line will contain two integers n, k (1 ≤ n ≤ 50, 1 ≤ k ≤ min{1018, l} where l is the length of the Kyoya's list).

Output
Print n space-separated integers, representing the permutation that is the answer for the question.

Examples
input
4 3
output
1 3 2 4
input
10 1
output
1 2 3 4 5 6 7 8 9 10
Note
The standard cycle representation is (1)(32)(4), which after removing parenthesis gives us the original permutation. The first permutation on the list would be [1, 2, 3, 4], while the second permutation would be [1, 2, 4, 3].

题意：

cycle，就是不在同一个位置的数，会成为一个圈

每个cycle从大到小排序，所有cycle按第一个元素从大到小排序

排序完了以后跟原来的序列不变就是合法

然后让你输出第k大的合法数组

思路：看卿神的吧

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
ll c[110],k;
int n,a[110];
int main(){
    c[0]=c[1]=1LL;
    for(int i=2;i<100;i++) c[i]=c[i-1]+c[i-2];
    scanf("%d%I64d",&n,&k);
    for(int i=1;i<=n;){
        if(k>c[n-i]){
            k-=c[n-i];
            a[i]=i+1;
            a[i+1]=i;
            i+=2;
        } else {
            a[i]=i;
            i++;
        }
    }
    for(int i=1;i<=n;i++) cout<<a[i]<<" ";
    cout<<endl;
    return 0;
}