hdu 1018 Big Number

Problem Description

In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.

 

Input

Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 10⁷ on each line.

 

Output

The output contains the number of digits in the factorial of the integers appearing in the input.

 

Sample Input

2 10 20

 

Sample Output

7 19

 

题目的意思是给出一个数，让你求这个数阶乘的位数。

网上的大神给出的方法是，将n!表示成10的次幂,即n!=10^M(10的M次方，10^2是3位M+1就代表位数)则不小于M的最小整数就是     n!的位数，对该式两边取对数，有M=log10^n!即：     M = log10^1+log10^2+log10^3...+log10^n     循环求和,就能算得M值，该M是n!的精确位数。

 

#include <cstdio>#include <cmath>int main(){int i, t, n;double sum;scanf("%d", &t);while (t--){sum = 1;scanf("%d", &n);for (i = 1; i <= n; i++)sum += log10(i);printf("%d/n", (int)sum);}return 0;}当数比较大的时候，可以利用斯特林公式：#include#includeconst double PI = 3.141592653589;int main() {int n, t;double sum;scanf("%d", &t);while (t--) {scanf("%d", &n);if (n == 0 || n == 1) { printf("1n"); continue; }sum = (n*log(n) - n + 0.5*log(2 * PI*n)) / log(10) + 1;printf("%dn", (int)sum);}return 0;}