HDU 1018 Big Number (簡單數學)

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Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25649    Accepted Submission(s): 11635

Problem DescriptionIn 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. 

 

InputInput 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. 

 

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

 

Sample Input21020 

 

Sample Output719 

 

SourceAsia 2002, Dhaka (Bengal) 

 

RecommendJGShining 

 

這道題我一開始用的大數求階乘的方法做的，結果逾時，O(N2)的演算法傷不起。

先掛一下逾時的代碼

 1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 int main() 7 { 8     int kase,num[3000],i,j; 9     scanf("%d",&kase);10     while(kase--)11     {12         int n;13         scanf("%d",&n);14         memset(num,0,sizeof(num));15         num[0]=1;16         for(int i=2;i<=n;i++)17         {18             int c=0;19             for(int j=0;j<3000;j++)20             {21                 num[j]=num[j]*i+c;22                 c=num[j]/10;23                 num[j]=num[j]%10;24             }25         }26         for(i=2999;i>=0;i--)27             if(num[i])28                 break;29         printf("%d\n",i+1);30     }31     return 0;32 }

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後來一想，就算不逾時，10^7的階乘也存不下，所以一時之間沒有思路。

後來上網看大神怎麼做的，才AC了。

如果要計算一個數num的位元，那麼可以用到 (int)lg((double)num)+1

這裡lg(1*2*3......n)=lg1+lg2+lg3+......lg n

這裡的話可以將1~10^7的數的位元全部存起來，是一種打表的做法。

 1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 #include<stdlib.h> 5 using namespace std; 6 int a[10000005]; 7 int main() 8 { 9     a[1]=1;10     double sum=0;11     for(int i=2;i<=10000000;i++)12     {13         sum+=log10((double)i);14         a[i]=sum+1;15     }16     int kase,n;17     scanf("%d",&kase);18     while(kase--)19     {20         scanf("%d",&n);21         printf("%d\n",a[n]);22     }23     return 0;24 }

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當然還有一種做法就是用到了斯特林數

公式為：

求出階乘然後同樣的方法取位元,求出lg(n!)再向上取整

 1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 #include<stdlib.h> 5 const double PI=3.141592654; 6 const double e=2.718281828; 7 using namespace std; 8 int main() 9 {10     int kase,n;11     double sum;12     scanf("%d",&kase);13     while(kase--)14     {15         scanf("%d",&n);16         sum=log10(2*PI*n)/2+n*log10(n/e);17         printf("%d\n",(int)sum+1);18     }19     return 0;20 }

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