hdu1018

2017-11-23

hdu1018 Big Number

hdu 1018 Big Number

Problem 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 ≤ n ≤ 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

Q:如何求一个数字有多少位？

A:用以10为底的对数解决
所以是[lgN]+1

#include<stdio.h>
#include<math.h> 
int main()
{
    int n;
    scanf("%d",&n);
    printf("%d",(int)log10(n)+1);
    return 0;
}

Q:如何解决阶乘？

A:两侧同时取对数,使用对数运算公式，lgn!=lg1+lg2+lg3+…+lg(n-1)+lgn

#include<stdio.h>
#include<math.h>
int main()
{
    int n,i,t;
    double sum;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&i);
        sum=0;
        for(t=1;t<=i;t++)
            sum+=log10(t);
        printf("%d\n",(int)sum+1);
    }
    return 0;
}

更简便的做法，使用斯特林(Stirling)公式解决n!的计算

斯特林公式(Stirling’s approximation)

#include<stdio.h>
#include<math.h>
#define pi 3.1415926
#define e 2.718281
int main()
{
    int n,i;
    double res;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&i);
        res=0.5*log10(2*pi*i)+i*log10(i/e);
        printf("%d\n",(int)res+1);
    }
    return 0;
}