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Posted by 15914304086 at 2012-08-01 09:13:06 on Problem 1423 and last updated at 2012-08-01 09:13:40
In Reply To:其实这道题可以用这个公式 Posted by:slp at 2010-09-14 22:34:35
对于n!,其位数为k的话,那么n!等于a*10^k(0<a<1),即
n!=a*10^k,两边取对数
log(n!)=log(a)+klog(10)=log(a)+k....................(以10为底取对数)
log(n!)=log(n)+log(n-1)+log(n-2)+....+log(3)+log(2)+log(1)
这时候要注意,a的范围其实进一步是0.1<a<1,所以-1<log(a)<0,
那么
k=-log(a)+log(n)+log(n-1)+...+log(2)+log(1),k是整数
取log(a)约为1,就是上面的结果。

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