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我的推导过程

Posted by xuguangshengqq at 2007-07-19 22:05:39 on Problem 1183
    1/a = (1/b+1/c) / (1 - (1/b) * (1/c))
   1/a = (b+c)/bc / (1-1/bc)
     a = (bc-1) / (b+c)
	 令 m = b+c
	 则 c = m - b;
	    m = (b^2+1)/(b-a)
	 令 i = b-a
	 则 m = i + 2a + (a^2+1)/i    //最后表达式
	    i+ (a^2+1)/i >= 2*sqrt(a^2+1)
		所以: i >= sqrt(a^2+1) 或 i<= sqrt(a^2+1)
	 则 m >= 2*sqrt(a^2+1) + 2a
     所以可以令 i 从 a 开始循环 i--;
	 直到 (a^2+1)%i == 0 再求出m即可


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