POJ Problem 1423 Big Number 【stirling公式】

Big Number

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27709		Accepted: 8837

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 <= m <= 10^7 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

Source

Dhaka 2002

这道题考察stirling公式。当n充分大时，n！与sqrt（2πn * pow（n/e,n））十分接近。利用此公式，可以对阶乘有更好的估算。

，对于这道题，两边同时取以十为底的对数后，n！的位数就等于右边式子加上一。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <iostream>
#define MAX_N   100005
#define MAX(a, b) (a > b)? a: b
#define MIN(a, b) (a < b)? a: b
using namespace std;
const double e = 2.71828182845904523536;
const double PI = 3.141592653589793238;

double strling(int n) {
    return 0.5*log10(2*PI*n) + n*log10(n/e);
}
int main() {
    int t, m;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &m);
        printf("%d
", (int)strling(m) + 1);
    }
    return 0;
}