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*n)=log10(1)+log10(2)+…求位数：log10(n!)=log10(1*2*3…+log10(n)，对log10(n!)的值取整加1就是n!的位数。忽略log10(1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3)) ≈ log10(1) = 0.) = log10(sqrt(2 * pi * n)) + n * log10(n / e)#include<iostream>#include<cmath>using namespace std;len = (int)(log10(sqrt(2 * Pi * n)) + n *... 阅70 转0 评0 公众公开 13-03-13 21:25 |
