HDU 1018 Big Number（求n！的位数）

转载请注明出处：忆梦http://blog.csdn.net/yimeng2013/article/details/12856081

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1018

题意：计算n！结果的位数（1<= n <= 10^7）
题解：
①我们先来讨论任意一个数a如何求解其位数,
设a的位数为x，既有：10^（x-1） <= a < 10^x
取对数有：log10(10^(x-1)) <= log10(a) < log10(10^x)
即:x-1 <= log10(a) < x;
则:(int)log10(a) = x-1;
所以: x =
(int)log10(a)+1;
②n! = 1*2*3……*n
所以[b]n! 的位数 = (int)log10(1*2*3……*n)+1 = (int)(log10(1)+log10(2)+……+log10(n))
+ 1;[/b]

超内存的代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#define N 10000000
int ans[N+5];
int main ()
{
double s = 0;
for(int i = 1; i <= N; i++)
{
s += log10(i);
ans[i] = (int)s + 1;
}

int T;
scanf("%d", &T);
while(T--)
{
int n;
scanf("%d", &n);
printf("%d\n", ans
);
}
return 0;
}

AC：968MS

#include<cstdio>
#include<cmath>

int main ()
{

int T;
scanf("%d", &T);
while(T--)
{
int n;
scanf("%d", &n);

double s = 0;
for(int i = 1; i <= n; i++)
s += log10(i);

printf("%d\n", int(s)+1);
}
return 0;
}

网上0MS的程序

《计算机程序设计艺术》中给出了另一个公式
 n! = sqrt(2*π*n) * ((n/e)^n) * (1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3))

 π = acos(-1)

 e = exp(1)


两边对10取对数

忽略log10(1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3)) ≈ log10(1) = 0

得到公式

    log10(n!) = log10(sqrt(2 * pi * n)) + n * log10(n / e)

#include<iostream>
#include<cmath>
using namespace std;

int main()
{
double Pi = acos(-1.0);
double E = exp(1.0);
int n,m;
cin >> m;
while(m--)
{
cin >> n;
double len = (log10(sqrt(2 * Pi * n)) + n * log10(n/E) );
cout <<(int) len +1 << endl;
}
return 0;
}