poj1423 hdu1018 Big Number

Big Number

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 25225 Accepted: 8079

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 <= m <= 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

21020

Sample Output

719

在杭电上交过了但是poj上便tle了

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

下面这个在杭电和poj上全过了

斯特林公式

#include<stdio.h>#include<math.h>#define  e 2.718281828459045#define pi 3.1415926int main(){    int m,n,l;    scanf("%d",&n);    while(n--)    {        scanf("%d",&m);        printf("%d\n",(int)(0.5*log10(2*pi*m)+m*log10(m/e))+1);    }    return 0;}

被坑了好多次,这道题开始以为大数题目后拉总是错误于是就上网看了一下别人的代码发现有特殊做法便是用斯特林公式

但是因为+1在括号内导致当m=1时答案是0wa了好多次......