hdu 1018 Big Number (数学题)
Problem Description
Inmany applications very large integers numbers are required. Some of theseapplications are using keys for secure transmission of data, encryption, etc.In this problem you are given a number, you have to determine the number ofdigits in the
factorial of the number.
Input
Inputconsists of several lines of integer numbers. The first line contains aninteger n, which is the number of cases to be tested, followed by n lines, oneinteger 1 ≤ n ≤ 107 on each line.
Output
Theoutput contains the number of digits in the factorial of the integers appearingin the input.
SampleInput
2
10
20
Sample Output
7
19
/**************************************************
// 不能直接算N!,数据规模 1<N<10^7 太大,超出 2^31 的范围,所以取对数函数
// N = M*10^n n = log10(N) log10() 函数在头文件cmath中
984 MS,差点超时
************************************************/
#include <iostream>
#include<cmath>
using namespace std;
int main()
{
double sum;
int T,n;
cin>>T;
while(T--)
{
cin>>n;
sum = 1;
for(int i = 1;i<=n;i++)
sum+=log10(i);
cout<<(int)sum<<endl;
}
return 0;
}
hdu 1018 Big Number (数学题)的更多相关文章
- HDU 1018 Big Number
LINK:HDU 1018 题意:求n!的位数~ 由于n!最后得到的数是十进制,故对于一个十进制数,求其位数可以对该数取其10的对数,最后再加1~ 易知:n!=n*(n-1)*(n-2)*...... ...
- HDU 1018 Big Number (数学题)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1018 解题报告:输入一个n,求n!有多少位. 首先任意一个数 x 的位数 = (int)log10(x ...
- hdu 1018:Big Number(水题)
Big Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...
- hdu 1018 Big Number 数学结论
Big Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...
- HDU 1018 Big Number【斯特林公式/log10 / N!】
Big Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...
- HDU 1018 Big Number (log函数求数的位数)
Problem Description In many applications very large integers numbers are required. Some of these app ...
- HDU 1018 Big Number 数学题解
Problem Description In many applications very large integers numbers are required. Some of these app ...
- HDU 1018 Big Number 斯特林公式
Big Number 题意:算n!的位数. 题解:对于一个数来算位数我们一般都是用while去进行计算,但是n!这个数太大了,我们做不到先算出来在去用while算位数. while(a){ cnt++ ...
- HDU 1018 Big Number (阶乘位数)
题意: 给一个数n,返回该数的阶乘结果是一个多少位(十进制位)的整数. 思路: 用对数log来实现. 举个例子 一个三位数n 满足102 <= n < 103: 那么它的位数w 满足 w ...
随机推荐
- Bzoj 1579: [Usaco2009 Feb]Revamping Trails 道路升级 dijkstra,堆,分层图
1579: [Usaco2009 Feb]Revamping Trails 道路升级 Time Limit: 10 Sec Memory Limit: 64 MBSubmit: 1573 Solv ...
- Python的模块,模块的使用、安装,别名,作用域等概念
所谓的模块就是将不同功能的函数分别放到不同的文件中,这样不仅有利于函数的维护,也方便了函数的调用.在Python中,一个.py文件就是一个模块(Module). 在模块的上层有一个叫做包(Packag ...
- Using Live555 to Stream Live Video from an IP camera connected to an H264 encoder
http://stackoverflow.com/questions/27279161/using-live555-to-stream-live-video-from-an-ip-camera-con ...
- kvm usb2.0
Virt-Manager adds support for usb2 Wednesday, April 4, 2012 - 10:40 Haydn Solomon The most recent re ...
- Quartz定时任务学习(二)web应用
web中使用Quartz 1.首先在web.xml文件中加入 如下内容(根据自己情况设定) 在web.xml中添加QuartzInitializerServlet,Quartz为能够在web应用中使用 ...
- Codeforces 114A-Cifera(暴力)
A. Cifera time limit per test 2 seconds memory limit per test 256 megabytes input standard input out ...
- SSH服务
基于Linux的服务器有多个网卡,其中一个网卡连接了网线,通过该网线链接了个人PC.PC上启动Vmware虚拟机,启动ubuntu系统.然后设置PC的网络为自动获取IP,在PC的Linux的Ubunt ...
- Memcached笔记——(四)应对高并发攻击【转】
http://snowolf.iteye.com/blog/1677495 近半个月过得很痛苦,主要是产品上线后,引来无数机器用户恶意攻击,不停的刷新产品各个服务入口,制造垃圾数据,消耗资源.他们的最 ...
- 四种可变交流swap方法
1.void swap(int &x, int &y){ int temp=x; x=y; y=temp; } 2.void swap(int &x, int &y){ ...
- hdu1864 最大报销额(01背包)
转载请注明出处:http://blog.csdn.net/u012860063 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1864 Problem ...