Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31053 Accepted Submission(s): 14398
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
Sample Output
7
19
一个整数n的位数：log10(n)+1
此题答案为：log10(n!)+1.
log(n!)=log(n)+log(n-1)+log(n-2).....+log(1)
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
int main()
{
int i,j,n,T,res;
double t;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
t=0;
for(i=1;i<=n;i++)
t+=log10(i);
res=(int)t+1;
printf("%d\n",res);
}
return 0;
}
