HDU 5916 Harmonic Value Description 【构造】(2016中国大学生程序设计竞赛(长春))

Harmonic Value Description

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0
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

 

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5916

题目大意：

　　给你N和K（1<=K<2K<=2N），求一个1~2N的排列，使得∑gcd(A_i,A_i+1)恰为第K小（等视为相同小）。

题目思路：

　　【构造】

　　这题一看2K<=2N，可能就是构造题。如果1,2,3...2N排列的话∑gcd(A_i,A_i+1)为第1小。

　　若要∑gcd(A_i,A_i+1)=K，只需要将2K和K提出来，将2K放在开头，K放在第二个，接下来只要满足相邻两个数仍然相差1即可。

　　只要将K-1~1接在K后面，把K+1~2N接在1后面，去掉2K即可。（2K-1，2K+1为奇数，gcd=1）

//
//by coolxxx
//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<iomanip>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<memory.h>
#include<time.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//#include<stdbool.h>
#include<math.h>
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define abs(a) ((a)>0?(a):(-(a)))
#define lowbit(a) (a&(-a))
#define sqr(a) ((a)*(a))
#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
#define mem(a,b) memset(a,b,sizeof(a))
#define eps (1e-10)
#define J 10000
#define mod 1000000007
#define MAX 0x7f7f7f7f
#define PI 3.14159265358979323
#pragma comment(linker,"/STACK:1024000000,1024000000")
#define N 104
using namespace std;
typedef long long LL;
double anss;
LL aans,sum;
int cas,cass;
int n,m,lll,ans;
int main()
{
    #ifndef ONLINE_JUDGEW
//    freopen("1.txt","r",stdin);
//    freopen("2.txt","w",stdout);
    #endif
    int i,j,k;
//    init();
//    for(scanf("%d",&cass);cass;cass--)
    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
//    while(~scanf("%s",s))
//    while(~scanf("%d",&n))
    {
        scanf("%d%d",&n,&m);
        printf("Case #%d: ",cass);
        printf("%d %d",m+m,m);
        for(i=m-1;i;i--)printf(" %d",i);
        for(i=m+1;i<=n;i++)
            if(i!=m+m)printf(" %d",i);
        puts("");
    }
    return 0;
}
/*
//

//
*/