SGU491 Game for Little Johnny

491

Accepted

3522 ms

27367 kb

题目大意：给一个N。让你求有多少对A和B，让方程Ax+By=N有自然数解。（N<=100000）

分析：先枚举a再枚举x。对于每个a算出所有可能的b，然后对于b去重。最后便可得到所有的（a，b）对个数。

参考博客：WJBZBMR的空间

 

  
    

   
     
   
     /*
   
     total 100tests 3522 ms
   
     */
   
     
#include
   
     <
   
     iostream
   
     >
   
     
#include
   
     <
   
     algorithm
   
     >
   
     
#include
   
     <
   
     vector
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;

   
     const
   
      
   
     int
   
      Max
   
     =
   
     100001
   
     ;
vector
   
     <
   
     int
   
     >
   
      factor[Max]; 
   
     //
   
     A[i]里面按升序存的i的所有约数
   
     

   
     typedef vector
   
     <
   
     int
   
     >
   
     ::iterator it;
typedef vector
   
     <
   
     int
   
     >
   
     ::reverse_iterator rit;

   
     int
   
      enumb[Max
   
     *
   
     50
   
     ]; 
   
     //
   
     对于给定的a，枚举所有的b
   
     

   
     int
   
      n;

   
     void
   
      cal(
   
     int
   
      x)
{
 
   
     int
   
      tmp[
   
     10000
   
     ];
 
   
     int
   
      cnt
   
     =
   
     0
   
     ;
 
   
     for
   
     (
   
     int
   
      i
   
     =
   
     1
   
     ;i
   
     *
   
     i
   
     <=
   
     x;i
   
     ++
   
     )
 
   
     if
   
     (x
   
     %
   
     i
   
     ==
   
     0
   
     )
 tmp[cnt
   
     ++
   
     ]
   
     =
   
     i,tmp[cnt
   
     ++
   
     ]
   
     =
   
     x
   
     /
   
     i;
 sort(tmp,tmp
   
     +
   
     cnt);
 factor[x].push_back(tmp[
   
     0
   
     ]);
 
   
     for
   
     (
   
     int
   
      i
   
     =
   
     1
   
     ;i
   
     <
   
     cnt;i
   
     ++
   
     )
 
   
     if
   
     (tmp[i]
   
     !=
   
     tmp[i
   
     -
   
     1
   
     ])
 factor[x].push_back(tmp[i]);
}


   
     int
   
      main()
{

   
     //
   
      freopen("in.txt","r",stdin);
   
     

   
      
   
     for
   
     (
   
     int
   
      i
   
     =
   
     1
   
     ;i
   
     <
   
     Max;i
   
     ++
   
     )
 cal(i);
 
   
     while
   
     (cin
   
     >>
   
     n)
 {
 
   
     int
   
      ans
   
     =
   
     0
   
     ;
 
   
     for
   
     (
   
     int
   
      a
   
     =
   
     1
   
     ;a
   
     <
   
     n;a
   
     ++
   
     )
 {
 
   
     int
   
      
   
     *
   
     end
   
     =
   
     enumb;
 
   
     for
   
     (
   
     int
   
      t
   
     =
   
     a;t
   
     <
   
     n;t
   
     +=
   
     a)
 {
 
   
     int
   
      res
   
     =
   
     n
   
     -
   
     t;
 
   
     for
   
     (rit i
   
     =
   
     factor[res].rbegin();i
   
     !=
   
     factor[res].rend();i
   
     ++
   
     )
 {
 
   
     if
   
     (
   
     *
   
     i
   
     >
   
     a) 
   
     *
   
     (end
   
     ++
   
     )
   
     =*
   
     i;
 
   
     else
   
      
   
     break
   
     ;
 }
 }
 
   
     if
   
     (enumb
   
     ==
   
     end) 
   
     continue
   
     ;
 sort(enumb,end);
 ans
   
     ++
   
     ;
 
   
     for
   
     (
   
     int
   
      
   
     *
   
     i
   
     =
   
     enumb
   
     +
   
     1
   
     ;i
   
     !=
   
     end;i
   
     ++
   
     )
 
   
     if
   
     (
   
     *
   
     i
   
     !=*
   
     (i
   
     -
   
     1
   
     )) ans
   
     ++
   
     ;
 }
 cout
   
     <<
   
     ans
   
     <<
   
     endl;
 }
 
   
     return
   
      
   
     0
   
     ;
}