HUT-1602 GCD depth

1602: GCD depth

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 24   Solved: 3
[ Submit][ Status][ Web Board]

Description

In mathematics, the greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder. For example, the GCD of 8 and 12 is 4.
------ From Wikipedia

    In this problem,we will introduce a new value related to GCD,which is GCD depth. To GCD depth,follow function will discribe it clearly.

    int GCD_depth( int x , int y ) {
        if ( y == 0 ) return 0;
        else return GCD_depth( y , x%y ) + 1; 
    }

    And we define the GCD depth of y with x is GCD_depth(x,y).For example , GCD depth of 5 with 3 is 4.You can find the GCD depth of two numbers easily ,but LH wants know that: for a number x, how many numbers meet the condition that the GCD depth with x equals to d in the interval [y0,y1]? So please help LH to find the answer quickly.

Input

  There are several test cases, each test case contains four Non-negative integers x( 0 <= x <= 200000) , d( 0 <= d <= 30 ),y0 ,y1(0 <= y0 <= y1 <= 10^9),which descripted as above. 
    The input will finish with the end of file.

Output

   For each the case, just output a integer which represent the number of integers meeted the discripted condition. 

Sample Input

7 2 0 5 3 0 0 1 11 1 2 8

Sample Output

2 1 0

 

　　四个数 N， M， S， E， 求 GCD_depth( N, i ) == M 的个数，其中 i 属于 [s, e]。

　　思路：该题我们只要将 i 属于 [0 - N) 的GCD_depth计算出来，然后大于N的数就用 GCD_depth( N, i% N ) + 2 == M 去统计了，自己推一下就能明白为什么了。 注意当这个数等于N的时候比较特殊，这时候是加1而不是加 2。 主要是区间的读取边界太难处理了，代码比较丑。

　　

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 using namespace std;
 6 
 7 int gcd( int a, int b )
 8 {
 9  if( b== 0 )
 10  {
 11  return 0;
 12  }
 13  else return gcd( b, a% b )+ 1;
 14 }
 15 
 16 int rec[200005];
 17 
 18 int main()
 19 {
 20  int N, M, s, e;
 21  while( scanf( "%d %d %d %d", &N, &M, &s, &e )!= EOF )
 22  {
 23  int base= 0, cnt= 0, beg, end;
 24  if( N== 0 )
 25  {
 26  if( M== 1 )
 27  {
 28  int k= 1;
 29  if( s== 0 )
 30  {
 31  k= 0;
 32  }
 33  cnt= e- s+ k;
 34  }
 35  else if( M== 0 )
 36  {
 37  int k= 0;
 38  if( s== 0 )
 39  {
 40  k= 1;
 41  }
 42  cnt= k;
 43  }
 44  printf( "%d\n", cnt );
 45  continue;
 46  }
 47  for( int i= 0; i< N; ++i )
 48  {
 49  rec[i]= gcd( N, i );
 50  if( rec[i]+ 2== M )
 51  {
 52  base++;
 53  }
 54  }
 55  if( s>= N )
 56  { 
 57  beg= ( int )ceil( 1.0* s/ N ), end= ( int )floor( 1.0* e/ N );
 58  for( int i= s; i< beg* N; ++i )
 59  {
 60  if( rec[i% N]+ 2== M )
 61  {
 62  cnt++;
 63  }
 64  }
 65  for( int i= end* N; i<= e; ++i )
 66  {
 67  if( i!= N )
 68  {
 69  if( rec[i% N]+ 2== M )
 70  {
 71  cnt++;
 72  }
 73  }
 74  }
 75  cnt+= base* ( end- beg );
 76  if( s== N )
 77  {
 78  if( rec[0]+ 1== M )
 79  {
 80  cnt++;
 81  }
 82  else if( rec[0]+ 2== M )
 83  {
 84  cnt--;
 85  }
 86  }
 87  }
 88  else if( e>= N )
 89  {
 90  end= ( int )floor( 1.0* e/ N );
 91  for( int i= s; i< N; ++i )
 92  {
 93  if( rec[i]== M )
 94  {
 95  cnt++;
 96  }
 97  }
 98  for( int i= end* N; i<= e; ++i )
 99  {
100  if( i!= N )
101  {
102  if( rec[i% N]+ 2== M )
103  {
104  cnt++;
105  }
106  }
107  }
108  cnt+= base* ( end- 1 );
109  if( rec[0]+ 1== M )
110  {
111  cnt++;
112  }
113  else if( rec[0]+ 2== M )
114  {
115  cnt--;
116  }
117  }
118  else
119  {
120  for( int i= s; i<= e; ++i )
121  {
122  if( rec[i]== M )
123  {
124  cnt++;
125  }
126  }
127  }
128  printf( "%d\n", cnt );
129  }
130 }