HDU_1018

题目链接

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方法1

$\because n!=1\times 2\times ...\times n$

$\therefore n$ 的位数

$=\lceil lg(n!)\rceil$

$=\lceil lg(1\times 2\times ...\times n)\rceil$

$=\lceil lg(1)+lg(2)+...+lg(n)\rceil$

注：

• $lg(x)=lo{g}_{10}x$
• $\lceil \rceil$ 表示向上取整

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define LL long long
#define mem(a, b) memset(a, b, sizeof a)
#define lowbit(x) (-x&x)
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define endl '\n'
#define rev(x) reverse(x.begin(), x.end())
#define INF 0x3f3f3f3f

using namespace std;

double ans;
int n;

void solve() {
	int tt;
    cin >> tt;
    // tt = 1;
    while (tt -- ) {
    	cin >> n;
    	ans = 0;
    	
    	for (int i = 1; i <= n; i ++ ) ans += log10(i);
    	
    	printf("%d\n", (int)(ans + 1));
    }
}

int main() {
	IOS;
	
	solve();
	
	return 0;
}
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方法二

利用斯特林公式($Stirlin{g}^{\prime }sapproximation$)。

$\because n!\approx \sqrt{2\pi n}{(\frac{n}{e})}^n$

$\therefore n$ 的位数

$=\lceil lg(n!)\rceil$

$=\lceil \frac{1}{2}lg(2\pi n)+nlg(\frac{n}{e})\rceil$

$=\lceil \frac{\frac{1}{2}ln(2\pi n)+nln(\frac{n}{e})}{ln(10)}\rceil$

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define LL long long
#define mem(a, b) memset(a, b, sizeof a)
#define lowbit(x) (-x&x)
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define endl '\n'
#define rev(x) reverse(x.begin(), x.end())
#define INF 0x3f3f3f3f
#define PI acos(-1)

using namespace std;

double ans;
int n;

void solve() {
	int tt;
    cin >> tt;
    // tt = 1;
    while (tt -- ) {
    	cin >> n;
    	
    	ans = ceil((0.5 * log(2 * PI * n) + n * log(n) - n) / log(10));
    	
    	printf("%d\n", (int)ans);
    }
}

int main() {
	IOS;
	
	solve();
	
	return 0;
}
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