Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 24637 | Accepted: 7895 |
Description
In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.
Input
Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 <= m <= 10^7 on each line.
Output
The output contains the number of digits in the factorial of the integers appearing in the input.
Sample Input
2 10 20
Sample Output
7 19
Source
思路:log10(n!)=log10(1*2*3…*n)=log10(1)+log10(2)+…+log10(n)
方法一:用空间换时间,开个数组把结果存起来一部分!
#include"stdio.h"
#include"math.h"
#include"string.h"
#define N 8000000
double a[N];
void Inti()
{
int i;
a[0]=0;
for(i=1;i<N;i++)
a[i]=a[i-1]+log10(i);
}
int main()
{
int t,n,i;
double s;
Inti();
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
if(n<N)
{
printf("%d\n",(int)a[n]+1);
continue;
}
s=a[N-1];
for(i=N;i<=n;i++)
s+=log10(i);
printf("%d\n",(int)ceil(s));
}
return 0;
}
方法二:
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)。
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"stdio.h"
#include"math.h"
#include"string.h"
#define N 80000
int main()
{
int t,n,i;
double s,pi,e;
pi=acos(-1);
e=exp(1);
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
s=log10(sqrt(2*pi*n))+n*log10(n/e);
if(n==1)
s=1;
printf("%d\n",(int)ceil(s));
}
return 0;
}