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Posted by Pzjay at 2009-11-04 10:51:21 on Problem 2853
In Reply To:Re:用笨办法太慢了，高手能指点指点吗? Posted by:chqch172443 at 2008-07-08 16:17:44 
假设num能分解成从j开始的连续i个数的和，则有i*j+i*(i-1)/2=num；
即i(2*j+i-1)=2*num,则j=(num-i*(i-1)/2)/i;j是大于0的整数，则(num-i*(i-1)/2)%i应为0，然后且满足j>=1即（num-i*(i-1)/2>=i,整理即得：i*(i+1)<=2*num;       ~over~
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