POJ 1942 Paths on a Grid

解题思路:组合数学

1->递推:C(m,n)=c(m-1,n)+C(m,n-1),时间复杂度O(m*n)

2->组合 ，时间复杂度O(min(m,n))

 

代码

   
     

    
      
    
      #include 
    
      <
    
      iostream
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;

    
      int
    
       main()
{
 __int64 i,m,n,ans;
 
    
      while
    
      (scanf(
    
      "
    
      %I64u %I64u
    
      "
    
      , 
    
      &
    
      m, 
    
      &
    
      n)
    
      &&
    
      m
    
      +
    
      n)
 {
 
    
      if
    
      (m
    
      >
    
      n)swap(m,n);
 
    
      for
    
      (ans
    
      =
    
      1
    
      ,i
    
      =
    
      0
    
      ;i
    
      <
    
      m;i
    
      ++
    
      )
 ans
    
      =
    
      ans
    
      *
    
      (m
    
      +
    
      n
    
      -
    
      i)
    
      /
    
      (i
    
      +
    
      1
    
      );
 printf(
    
      "
    
      %I64u\n
    
      "
    
      ,ans);
 }
 
    
      return
    
       
    
      0
    
      ;
}