HDU 1018 Big Number
Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38814 Accepted Submission(s): 18850
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
2
10
20
Sample Output
7
19
Source
Recommend
分析: 对于一个数n 它的位数等于 log10(n)+1;
那么n! 的位数为 log10(n*(n-1).....*1) +1
公式展开 log10(n)+log10(n-1)+....log10(1)+1;
代码如下:
#include <cstdio> #include <iostream> #include <cstring> #include <map> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; const int MAXN=40000; int ans[MAXN]; int flag; int main() { double sum; int t,n; scanf("%d",&t); while(t--) { sum=1; scanf("%d",&n); for(int i=1;i<=n;i++) { sum+=log10(i); } printf("%d\n",(int)sum); } return 0; }