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

长春] 6.Harmonic Value Description 规律+贪心"/>

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 [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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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[n];a[n]=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;
}