FZU1050 Number lengths(数论，规律，概念)

FZU1050 Number lengths(数论，规律，概念)

题目：

Accept: 1096    Submit: 2263
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

N! (N factorial) can be quite irritating and difficult to compute for large values of N. So instead of calculating N!, I want to know how many digits are in it. (Remember that N! = N * (N - 1) * (N - 2) * ... * 2 * 1)

 Input

Each line of the input will have a single integer N on it 0 < N < 1000000 (1 million). Input is terminated by end of file.

 Output

For each value of N, print out how many digits are in N!.

 Sample Input

 Sample Output

题目的意思是问你一个数的阶乘里面有几位数，以前做水题的时候做过，昨天比赛又遇到了，开一篇博客存一下记录阶乘位数的方法

求一个数阶乘的位数 
可以直接采用log10求一个数阶乘的位数 
N=n! 
方法1 
log10(n!)  
=log10(1∗2∗3…∗n)  
=log10(1)+log10(2)+…+log10(n)

log10N表示以10为底，N的对数。缩写是lgN.

natural logarithm 自然对数 
natural 英 [‘nætʃ(ə)r(ə)l] 美 [‘nætʃrəl] 
logarithm 英 [‘lɒgərɪð(ə)m; -rɪθ-] 美 [‘lɔɡərɪðəm]

代码：

#include <stdio.h>
#include <string.h>
#include <string>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define N 100+20
#define M 1000000+10
#define LL long long
using namespace std;
int a[M];
void biao()
{double x=1;for(int i=1; i<=M; i++){x+=log10(double(i));a[i]=floor(x);}}
int main()
{int n;biao();while(~scanf("%d",&n))printf("%dn",a[n]);return 0;
}