toj 2798Farey Sequence

Farey Sequence

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Time Limit: 3.0 Seconds    Memory Limit: 65536K    Multiple test files

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The Farey Sequence F_n for any integer n with n ≥ 2 is the set of irreducible rational numbers a/b with 0 < a < b ≤ n and gcd(a,b) = 1 arranged in increasing order. The first few are

F₂ = {1/2} F₃ = {1/3, 1/2, 2/3} F₄ = {1/4, 1/3, 1/2, 2/3, 3/4} F₅ = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5}

Now, your task is to print Farey Sequence given the value of n.

Input

There are several test cases. The first line is an integer giving the number of cases. Each test case has only one line, which contains a positive integer n.

Output

For each test case, you should output one line, which contains the corresponding Farey Sequence. Adjacent terms are separated by a single ',' and there can't be any white spaces in your output. See Sample Output for more clarifications on the output format.

Constraints

2 ≤ n ≤ 3000

Sample Input

4 2 3 4 5

Sample Output

1/2 1/3,1/2,2/3 1/4,1/3,1/2,2/3,3/4 1/5,1/4,1/3,2/5,1/2,3/5,2/3,3/4,4/5

Note

Do Not use cout to produce the output for this problem, since it is inefficient.

Source:  The 5th UESTC Programming Contest Problem ID in problemset: 2798

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#include  < iostream > #include  < queue > using   namespace  std; int  t,n; typedef  struct  node {      short  a,b;      float  c;     node(){}     node( short  aa, short  bb, float  cc)     {         a = aa;         b = bb;         c = cc;     }     friend  bool   operator   < (node x,node y)     {          return  x.c > y.c;     } }Point; priority_queue < Point > Q; Point p; int  gcd ( short  a ,  short  b) {    if  (b  ==   0 )        return  a;    return  gcd (b , a  %  b); } int  main() {          short  i,j;     scanf( " %d " , & t); //     int num = 0;      while (t -- )     {         scanf( " %d " , & n);         printf( " 1/%d " ,n);          while ( ! Q.empty())             Q.pop();          for (i = 1 ;i <= n - 1 ;i ++ )              for (j = i + 1 ;j <= n;j ++ )             {                  if (i == 1 && j == n)                      continue ;                  if (gcd(i,j) == 1 )                 {                     Q.push(node(i,j,i * 1.0 / j));                  //     num ++;                 }             }          while ( ! Q.empty())         {             p = Q.top();             Q.pop();             printf( " ,%d/%d " ,p.a,p.b);         }          // cout<<num<<endl;         printf( " \n " );     }      return   0 ; }

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/15/1457834.html