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Re:贴个公式在上面In Reply To:贴个公式在上面 Posted by:ccnu_xieshuo at 2008-04-29 19:20:18 > 设起始数为i，连续数的个数为j，则最后一个数为j+i-1;推得（2*i+j-1）*j=2*n，n就是那个要分的数；因为2*i-1>0；所以（2*i+j-1）*j=2*n>j*j,所以j<sqrt（2*n），对于j满足2*n%j=0&&(2*n/j-j+1)%2=0.一个j对应一个i，j就是要求的数目。是个o（n）的算法，不知道有没有更快的 Followed by:
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