Harmonic Value Description(构造题)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 786    Accepted Submission(s): 456
Special Judge

Problem Description

The harmonic value of the permutation p1,p2,⋯pn is

∑i=1n−1gcd(pi.pi+1)

Mr. Frog is wondering about the permutation whose harmonic value is the strictly k-th smallest among all the permutations of [n].

 

Input

The first line contains only one integer T (1≤T≤100), which indicates the number of test cases.

For each test case, there is only one line describing the given integers n and k (1≤2k≤n≤10000).

 

Output

For each test case, output one line “Case #x: p1 p2 ⋯ pn”, where x is the case number (starting from 1) and p1 p2 ⋯ pn is the answer.

 

Sample Input

2

4 1

4 2

 

Sample Output

Case #1: 4 1 3 2

Case #2: 2 4 1 3

//题目看上去比较复杂，其实就需要一点点思维，问 1-n n个数，第 k 小的harmonic value的排列是怎样的

题解: 要第 k 大就把前 k 个偶数丢前面去，剩下的不变就好了

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

int main()
{
    int T;
    cin>>T;
    for (int cnt=1;cnt<=T;cnt++)
    {
        int n ,k;
        scanf("%d%d",&n,&k);
        printf("Case #%d:",cnt);
        for (int i=1;i<=k;i++)
            printf(" %d",i*2);
        for (int i=1;i<=n;i++)
        {
            if (i%2==0&&i<=2*k) continue;
            printf(" %d",i);
        }
        printf("
");
    }
    return 0;
}