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HDU_1018

news 来源：原创 2024/11/27 17:58:18

题目链接

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方法1

$\because n!=1\times2\times...\times n$

$\therefore n$ 的位数

$\left\lceil lg(n!) \right\rceil$

$\left\lceil lg(1\times2\times...\times n) \right\rceil$

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

注：

$lg(x)=log_{10}x$
$\left\lceil \right\rceil$ 表示向上取整

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>

#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;
}

方法二

利用斯特林公式( $\ approximation$ )。

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

$\therefore \ n$ 的位数

$\left \lceil lg(n!) \right \rceil$

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

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

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>

#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;
}