CCPC[长春] 6.Harmonic Value Description 规律+贪心

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
.

 

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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经过总结规律，4个数的harmonic value为3 4....

5个的为 4 5 ....

6个的为5 6....

7个的为6 7 .....

8个的为7 8 ....

........

n个的为n-1  n ....

所以第k小的数的harmonic value为n-2+k

然后再利用贪心算法，把后面的数的位置放到适当位置，

使其harmonic value的值为n-2+k

具体分析看代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define PI acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef  long long ll;
typedef  unsigned long long  ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};

int T,n;
int a[10005];

int m,m1;
int main()
{

scanf("%d",&T);
for(int k=1;k<=T;k++)
{
memset(a,0,sizeof(a));
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++)
a[i]=i;
m--;
if(m%2==1)
{
for(int j=(m+1)*2;j>m+2;j--)//前移一个数
a[j]=a[j-1];
a[m+2]=(m+1)*2;
}
else
{
for(int j=m*2;j>m+1;j--)//前移一个数
a[j]=a[j-1];
a[m+1]=m*2;

if(n%2==0)//交换
{
int t;
t = a
;
a
=a[n-1];
a[n-1]=t;
}
else
{
int t;
t = a[n-3];
a[n-3]=a[n-2];
a[n-2]=t;
}
}
printf("Case #%d:",k);
for(int i=1;i<=n;i++)
printf(" %d",a[i]);
printf("\n");
}
return 0;
}