首页 > 代码库 > HDU5916

HDU5916

2024-08-16 00:54:06   208人阅读

Harmonic Value Description

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 287    Accepted Submission(s): 184
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
 

 

Source

2016中国大学生程序设计竞赛（长春）-重现赛
 
求gcd值第k小的1到n的排列。
 1 //2016.10.08 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5  6 using namespace std; 7  8 int main() 9 {10     int T, n, k, kase = 0;11     scanf("%d", &T);12     while(T--)13     {14         int tmp = -1;15         scanf("%d%d", &n, &k);16         printf("Case #%d:", ++kase);17         if(k == 1){18             for(int i = 1; i <= n; i++)19                   printf(" %d", i);20         }else{21             if(k&1){22                 printf(" %d %d", 2*k, k);23                 for(int i = 2; i <= n; i++)24                 {25                     if(i == 2*k)continue;26                     if(i == k)printf(" 1");27                     else printf(" %d", i);28                 }29             }else{30                 printf(" %d %d", 2*k, k);31                 for(int i = 1; i <= n; i++)32                 {33                     if(i == k || i == 2*k)continue;34                     else printf(" %d", i);35                 }36             }37         }38         printf("\n");39     }40 41     return 0;42 }

 

HDU5916


联系
我们