poj 2000 Gold Coins

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 13643		Accepted: 8495

  本题有一个小小的规律就是1*1+2*2+3*3+.........+（x-1）*（x-1）(x由（x+1)*x/2>n算出）有可能（x+1）*x/2!=n此时令s=(x+1)*x/2;sum=sum+(x-(s-n))*x

代码：

 

  
    

   
     
   
      1 
   
     #include
   
     <
   
     stdio.h
   
     >
   
     

   
      2 
   
      
   
     int
   
      main()

   
      3 
   
     {

   
      4 
   
      
   
     int
   
      n,s,i,sum,j;

   
      5 
   
      
   
     while
   
     (scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     n)
   
     !=
   
     EOF)

   
      6 
   
      {

   
      7 
   
      i
   
     =
   
     0
   
     ;s
   
     =
   
     0
   
     ;sum
   
     =
   
     0
   
     ;

   
      8 
   
      
   
     if
   
     (n
   
     ==
   
     0
   
     )

   
      9 
   
      
   
     break
   
     ;

   
     10 
   
      
   
     while
   
     (s
   
     <
   
     n)

   
     11 
   
      {

   
     12 
   
      i
   
     ++
   
     ;

   
     13 
   
      s
   
     =
   
     (i
   
     +
   
     1
   
     )
   
     *
   
     i
   
     /
   
     2
   
     ;

   
     14 
   
      }

   
     15 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <
   
     i;j
   
     ++
   
     )

   
     16 
   
      sum
   
     =
   
     sum
   
     +
   
     j
   
     *
   
     j;

   
     17 
   
      sum
   
     =
   
     sum
   
     +
   
     (i
   
     -
   
     s
   
     +
   
     n)
   
     *
   
     i;

   
     18 
   
      printf(
   
     "
   
     %d %d\n
   
     "
   
     ,n,sum);

   
     19 
   
      }

   
     20 
   
      
   
     return
   
      
   
     0
   
     ;

   
     21 
   
     }

   
     22