hdu1018——Big Number

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32419    Accepted Submission(s): 15180

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 ≤ 10⁷ on each line.

 

Output

The output contains the number of digits in the factorial of the integers appearing in the input.

 

Sample Input

21020

 

Sample Output

719

 

前两天写了N!的共精度实现，于是这题仍然妄想会直接想过，暴力算出N!，然后输出位数然。

于是。。。超时！

#include<iostream>#include<cstdio>using namespace std;long num[10000000];int main(){int n;scanf("%d",&n);while(n--){int a;scanf("%d",&a);long i,j,c,m=0;num[0]=1;for(i=1;i<=a;i++){c=0;for(j=0;j<=m;j++){num[j]=num[j]*i+c;c=num[j]/10;num[j]%=10;}while(c){num[++m]=c%10;c/=10;}}printf("%d\n",m+1);}return 0;}

之后想到了找规律的方法，分别找到了a的取10的对数，a开平方的数都没发现规律。也根据位数直接做了一些判断，还是不行。

晚上睡觉前，看了别人的博客，发现要走的路还很长！

求一个数n的位数，直接求n以10为底的对数不就好了吗？然后由于是阶乘，直接用公式化简，空间复杂度o（n）

通过这道题，还学到了另一个重要的东西——斯特林公式

然后直接用求对数的方法来求这个解；空间复杂度o（1)，有些东西简直不敢相信！

#include<iostream>#include<cmath>#define PI 3.1415926using namespace std;int main(){int n;cin>>n;while(n--){double a,res;cin>>a;res=(0.5*log(2*PI*a)+a*log(a)-a)/log(10)+1;cout<<(int)res<<endl;}return 0;} 

确实从开始的超时，到最后的o（1）让人难以置信，要走的路还很长！加油！