fzu 1050 Number lengths

题目链接：fzu 1050 Number lengths

题目大意：计算n!有多少位数。

解题思路：ans = log10(n!) =  ∑(1≤i≤n)log(i).

#include 
#include 

const int N = 1000005;
int n, d[N];

void init() {
	double sum = 0;
	for (int i = 1; i < N; i++) {
		sum += log10((double)i);
		d[i] = (int)sum;
	}
}

int main () {
	init();
	while (scanf("%d", &n) == 1) {
		printf("%d\n", d[n] + 1);
	}
	return 0;
}