求n^n和n！的最左边的数字

n^n 的位数 k = [lg(n^n)]+1=[n*lg(n)]+1;

最左边的数 x = n^n/10^(k-1);

取对数：lg(x) = n * lg(n) - (k-1) = (n*lg(n) - [n*lg(n)]);

最左边的数：[x] = [10^lg(x)] = [10^(n*lg(n) - [n*lg(n)])];

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
int main() {
    int n, t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        double a = n * log10(n);
        printf("%d
", (int)pow(10.0, a-(long long)a)); 
    }
}

Stirling公式:

根据Stirling公式n! 与√(2πn) * (n/e)^n的值十分接近 

n! 的位数 k = [lg(n!)]+1;

最左边的数 x = n!/10^(k-1);

取对数：lg(x) =  lg(n!) - (k-1) ;

lg(n!)  = lg(√(2πn) * (n/e)^n) = 1/2*(lg(2π) + lg(n)) + n * (lg(n) - lg(e));

代入[x] = [10^lg(x)] = [10^(lg(n!) - [lg(n!)])]即可。

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
const double PI = 3.1415926;
const double e = 2.718;
int main() {
    int n, t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        double b = (log10(2 * PI) + log10(n)) / 2 + n * (log10(n) - log10(e));  
        printf("%d
", (int)pow(10.0, b-(long long)b)); 
    }
}