toj_2800.cpp

Tianjin University Online Judge 的80多道题目 .

/*2800.   Cube Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 555   Accepted Runs: 298


As a student of the applied mathematics school of UESTC, WCM likes mathematics. Some day he found an interesting theorem that every positive integer's cube can be expressed as the sum of some continuous odd positive integers.

For example,

11*11*11 = 1331 = 111+113+115+117+119+121+123+125+127+129+131

Facing such a perfect theorem, WCM felt very agitated. But he didn't know how to prove it. He asked his good friend Tom Riddle for help. Tom Riddle is a student of the computer science school of UESTC and is skillful at programming. He used the computer to prove the theorem's validity easily. Can you also do it?

Given a positive integer N, you should determine how to express this number as the sum of N continuous odd positive integers. You only need to output the smallest and the largest number among the N integers.

Input
The input contains an integer on the first line, which indicates the number of test cases. Each test case contains one positive integer N on a single line(0 < N ≤ 1000).

Output
For each test case, output two integers on a line, the smallest and the largest number among the N continuous odd positive integers whose sum is N * N * N.
Sample Input

2
11
3

Sample Output

111 131
7 11



Source: The 5th UESTC Programming Contest*/
#include<cstdio>

int  getMid( int a , int b )
{
  int mid;
  mid = ( a + b ) / 2;
  if ( mid % 2  == 0 )
    return mid + 1;
  else
    return mid;
}

void find( int a ,  int& start , int mid ,  int& end , int cubeN ) 
{
  int result;
  result = ( a + end ) * ( end - a + 2 ) / 4;
  if ( start > mid && mid > end )
    {
      start = -1;
    }
  else  if ( result > cubeN )
    {
      start = mid;
      mid = getMid( start , end );
      find( a , start , mid , end , cubeN );
      printf( "start = %d , mid = %d , end = %d\n" , start , mid , end );
    }
  else if ( result < cubeN )
    {
      end = mid;
      mid = ( start + end ) / 2;
      mid = ( mid % 2 == 0 ) ? mid - 1 : mid;
      find( a , start , mid , end , cubeN );
      printf( "start = %d , mid = %d , end = %d\n" , start , mid , end );
    }
}

int main()
{
  int i , n , a , b , cubeN , start ,  mid ,  end , times;
  bool flag;
  scanf( "%d" , &times );
  for ( i = 0; i < times; i++ )
    {
      scanf( "%d" , &n );
      cubeN = n * n * n;
      a = 1;
      for ( a = 1; a < cubeN;  a+= 2 )
	{
	  start = a;
	  end = ( cubeN % 2 == 0 ) ? cubeN - 1: cubeN;
	  mid = getMid( start  , end );
	  find( a , start  , mid , end , cubeN );
	  if ( start != -1 )
	    {
	      b = end;
	      break;
	    }
	}
      printf( "%d %d\n" , a , b );
    }
  return 0;
}