标签：poj1423
Big Number
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 26151 |
|
Accepted: 8349 |
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
2
10
20
Sample Output
7
19
Source
#include <stdio.h>
#include <math.h>
const double PI = acos(-1.0);
const double E = exp(1.0);
int main() {
int t, n;
scanf("%d", &t);
while(t--) {
scanf("%d", &n);
printf("%d\n", (int)(log10(2*PI*n)/2+n*log10(n/E))+1);
}
return 0;
}
POJ1423 Big Number 【求N的阶乘的位数】
标签：poj1423
原文地址：http://blog.csdn.net/chang_mu/article/details/41575099
