杭电 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 ≤ 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

 

Source

Asia 2002, Dhaka (Bengal)

 

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题意：本题是要求阶乘的位数；

分析：此题看上去毫无头绪，但要是掌握了一个公式，就迎刃而解了；

x的位数就是（int）log10（x）+1；

推理如下：

对于任意一个给定的正整数a，
  假设10^(x-1)<=a<10^x，那么显然a的位数为x位，
  又因为
  log10(10^(x-1))<=log10(a)<(log10(10^x))
  即x-1<=log10(a)<x
  则(int)log10(a)=x-1,
  即(int)log10(a)+1=x
  即a的位数是(int)log10(a)+1

于是阶乘的位数便出来了，由于阶乘数比较大，故根据数学性质可写成循环实现；

代码如下：

 

int main(){int n,a,i,j;double s;scanf("%d",&n);for(i=0;i<n;i++){s=0;scanf("%d",&a);for(j=1;j<=a;j++)s+=log10(j);              //公式：x的位数就是（int）log10（x）+1;            printf("%d\n",(int)s+1); //故x阶乘的位数就是(int)log10(x!)+1;                }                         //因阶乘的位数较大，故x！用for循环来实现 return 0;}