hdoj 1018 Big Number

题目：

求整数n！的位数   1 <= n <= 10^7

类型：数论

有这样一个定理：f(n) = log10(n) + 1 f(n)为整数n的位数

证明：

设n的位数为m，则有10^(m - 1) <= n < 10^m，所以有m - 1 <= log10(n) < m

得：log10(n) < m <= log10(n) + 1

对log10(n)向下取整，得m = (int)log10(n) + 1，得证

所以有：f(n！) = f(1 * 2 * 3 * ... * n) + 1 = f(1)  + f(2) + f(3) +...f(n) + 1

// hdoj 1018 Big Number#include <iostream>#include <cmath>#include <cstdio>using namespace std;#define FOR(i,a,b) for(i = (a); i < (b); ++i)#define FORE(i,a,b) for(i = (a); i <= (b); ++i)int main(){    int i, j, n, m;    while(cin>>n) {        FOR(i, 0, n) {            cin>>m;            double end = 1;            FORE(j, 2, m)                end += log10(j * 1.0);            cout<<(int)end<<endl;        }    }    return 0;}