hdu 5916 Harmonic Value Description

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

24 14 2

 

Sample Output

Case #1: 4 1 3 2
Case #2: 2 4 1 3
构造序列。
一个正常序列相邻两数的gcd（）肯定为1.
很重要的一个条件，1<=2k<=n<=10000;
所以，提取出k 和 2k；
但是当k为奇数时。  gcd（k-1，k+1）可能不是1.
所以把1放在 k-1 和 k+1 之间 保证除了 k 可 2k ；其他gcd（）都为1。
#include<bits/stdc++.h>using namespace std;int main(){    int T,n,k,ca=1;    cin>>T;    while(T--)    {        scanf("%d%d",&n,&k);        printf("Case #%d: ",ca++);        if(n==1&&k==1) printf("1");        else            if(k&1)            {                printf("%d %d",2*k,k);                for(int i=2;i<k;i++)                    printf(" %d",i);                if(k!=1) printf(" 1");                for(int i=k+1;i<2*k;i++)                    printf(" %d",i);                for(int i=2*k+1;i<=n;i++)                    printf(" %d",i);            }            else            {                printf("%d %d",2*k,k);                for(int i=1;i<k;i++)                    printf(" %d",i);                for(int i=k+1;i<2*k;i++)                    printf(" %d",i);                for(int i=2*k+1;i<=n;i++)                    printf(" %d",i);            }        printf("\n");    }    return 0;}