标签：
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 JudgeProblem DescriptionThe 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].
InputThe 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).
OutputFor 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 Input2 4 1 4 2
Sample OutputCase #1: 4 1 3 2 Case #2: 2 4 1 3
Statistic | Submit | Clarifications | Back
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(Ai,Ai+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(Ai,Ai+1)为第1小。
若要∑gcd(Ai,Ai+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）
1 // 2 //by coolxxx 3 //#include<bits/stdc++.h> 4 #include<iostream> 5 #include<algorithm> 6 #include<string> 7 #include<iomanip> 8 #include<map> 9 #include<stack> 10 #include<queue> 11 #include<set> 12 #include<bitset> 13 #include<memory.h> 14 #include<time.h> 15 #include<stdio.h> 16 #include<stdlib.h> 17 #include<string.h> 18 //#include<stdbool.h> 19 #include<math.h> 20 #define min(a,b) ((a)<(b)?(a):(b)) 21 #define max(a,b) ((a)>(b)?(a):(b)) 22 #define abs(a) ((a)>0?(a):(-(a))) 23 #define lowbit(a) (a&(-a)) 24 #define sqr(a) ((a)*(a)) 25 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b)) 26 #define mem(a,b) memset(a,b,sizeof(a)) 27 #define eps (1e-10) 28 #define J 10000 29 #define mod 1000000007 30 #define MAX 0x7f7f7f7f 31 #define PI 3.14159265358979323 32 #pragma comment(linker,"/STACK:1024000000,1024000000") 33 #define N 104 34 using namespace std; 35 typedef long long LL; 36 double anss; 37 LL aans,sum; 38 int cas,cass; 39 int n,m,lll,ans; 40 int main() 41 { 42 #ifndef ONLINE_JUDGEW 43 // freopen("1.txt","r",stdin); 44 // freopen("2.txt","w",stdout); 45 #endif 46 int i,j,k; 47 // init(); 48 // for(scanf("%d",&cass);cass;cass--) 49 for(scanf("%d",&cas),cass=1;cass<=cas;cass++) 50 // while(~scanf("%s",s)) 51 // while(~scanf("%d",&n)) 52 { 53 scanf("%d%d",&n,&m); 54 printf("Case #%d: ",cass); 55 printf("%d %d",m+m,m); 56 for(i=m-1;i;i--)printf(" %d",i); 57 for(i=m+1;i<=n;i++) 58 if(i!=m+m)printf(" %d",i); 59 puts(""); 60 } 61 return 0; 62 } 63 /* 64 // 65 66 // 67 */
HDU 5916 Harmonic Value Description 【构造】(2016中国大学生程序设计竞赛(长春))
标签：
原文地址：http://www.cnblogs.com/Coolxxx/p/5930225.html
