ACMSTEP 2.3.3 Big Number //阶乘位数

原题链接


本题就是输入n 求 n! 有几位。
数学原理就是
如 1000！的位数就是 log10(1)+log10(2)+log10(3)+ ... + log10(999)+log10(1000) 后的值 向下取整 后 再 加 1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;
int main()
{
	int n, i, cases;
	double temp,resultn;
	cin>>cases;
	while (cases--)
	{
		cin>>n;
		resultn = 0;
		for (i=1; i<=n; i++)
		{
			temp = log10((double)i);
			resultn += temp;
		}
		resultn = floor(resultn) + 1;
	
		cout<<(int)resultn<<endl;
	}
	return 0;
}