最近一堆題目要補，一直咸魚，補了一堆水題都沒必要寫題解。備忘一下這個公式。

?Stirling公式的意義在于：當n足夠大時，n!計算起來十分困難，雖然有很多關于n!的等式，但并不能很好地對階乘結果進行估計，尤其是n很大之后，誤差將會非常大。但利用Stirling公式可以將階乘轉化成冪函數，使得階乘的結果得以更好的估計。而且n越大，估計得越準確。

傳送門:_(:з」∠)_?

再來一個詳細一點的，傳送門:( ･′ω`･ )?

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Big Number

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40645????Accepted Submission(s): 19863

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.

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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.

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

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Sample Input 2 10 20

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Sample Output 7 19

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Source Asia 2002, Dhaka (Bengal) 代碼: 1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 const double e=exp(1); 8 const double pi=acos(-1.0); 9 int main(){ 10 int t; 11 scanf("%d",&t); 12 while(t--){ 13 int n; 14 scanf("%d",&n); 15 if(n==1){printf("1\n");continue;} 16 double s=log10(2.0*pi)/2.0+(n+0.5)*log10(n)-n*log10(e); 17 int ans=ceil(s); 18 printf("%d\n",ans); 19 } 20 return 0; 21 }

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轉載于:https://www.cnblogs.com/ZERO-/p/8440970.html

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