求n^n和n！的最左边的数字

Posted 2020-10-15 白丁一枚

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\n", (int)pow(10.0, a-(long long)a)); 
    }
}