HUT1602 GCD depth

1602: GCD depth

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 24 Solved: 3
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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 }