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long long就够了key:输入一个数字是A=(n+1)^k，一个是B=n^k，必然一个偶数一个基数。分析偶数中2的个数，设为T，k必然是T的一个因数。而k的最大范围是31。最多做30次分析就够了，然后注意n=1的特殊情况。最后输出的2个数字中，一个小于B而另外一个不会超过30*A。so，只需要long long。 Followed by:
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