标签：each numbers test appear some tran lin cto nsis
传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1018
Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42715 Accepted Submission(s): 20844
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
Source
分析：
问你一个数的阶乘是多少位
n比较大
是大数问题
一开始不知道写
看了一下大佬的博客
具体分析；
一个正整数a的位数等于(int)log10(a) + 1
假设A=n!=1*2*3*......*n，那么我们要求的就是
(int)log10(A)+1，而：
log10(A)
=log10(1*2*3*......n) （根据log10(a*b) = log10(a) + log10(b)有）
=log10(1)+log10(2)+log10(3)+......+log10(n)
code：
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define max_v 105
int main()
{
/*
一个正整数a的位数等于(int)log10(a) + 1
假设A=n!=1*2*3*......*n，那么我们要求的就是
(int)log10(A)+1，而：
log10(A)
=log10(1*2*3*......n) （根据log10(a*b) = log10(a) + log10(b)有）
=log10(1)+log10(2)+log10(3)+......+log10(n)
*/
int n,t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
double sum=0;
for(int i=1; i<=n; i++)
{
sum+=log10((double)i);
}
printf("%d\n",(int)sum+1);
}
return 0;
}
HDU 1018Big Number（大数的阶乘的位数，利用公式）
标签：each numbers test appear some tran lin cto nsis
原文地址：https://www.cnblogs.com/yinbiao/p/9313368.html
