51nod N的阶乘

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输入N求N的阶乘的10进制表示的长度。例如6! = 720，长度为3。Input第1行：一个数T，表示后面用作输入测试的数的数量。（1 <= T <= 1000)第2 - T + 1行：每行1个数N。（1 <= N <= 10^9)Output共T行，输出对应的阶乘的长度。Input示例3456Output示例233

没什么好说的，斯特林近似公式。N=1的时候特判一下。不过没判也能够，应该是没有这组数据。下面是计算N！的位数的公式：

这里写图片描述

#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cctype>#include<cmath>#include<ctime>#include<string>#include<stack>#include<deque>#include<queue>#include<list>#include<set>#include<map>#include<cstdio>#include<limits.h>#define MOD 1000000007#define fir first#define sec second#define fin freopen("/home/ostreambaba/文档/input.txt", "r", stdin)#define fout freopen("/home/ostreambaba/文档/output.txt", "w", stdout)#define mes(x, m) memset(x, m, sizeof(x))#define Pii pair<int, int>#define Pll pair<ll, ll>#define INF 1e9+7#define Pi 4.0*atan(1.0)#define lowbit(x) (x&(-x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long ll;typedef unsigned long long ull;const double eps = 1e-12;//const int maxn = 1000100;using namespace std;#include<stdio.h>#include<algorithm>#include<iostream>#include<string.h>#include<math.h>using namespace std;inline int read(){    int x(0),f(1);    char ch=getchar();    while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();    return x*f;}ll Stirling(int n){    if(1 == n){        return 1;    }    ll res = log10(n*2*Pi)/2 + (ll)n*log10(n/exp(1.0)) + 1;    return res;}int main(){    int T, n;    T = read();    while(T--){        n = read();           printf("%lld\n", Stirling(n));    }    return 0;}

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