FZU1050 Number lengths(数论，规律，概念)

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题目：

Problem 1050 Number lengths

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

13 32000 

Sample Output

1 1 130271 

Submit Back Status Discuss

思路：

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 longusing 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("%d\n",a[n]);    return 0;}

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