hdu 1042 N!
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=1042
N!
Description
Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!
Input
One N in one line, process to the end of file.
Output
For each N, output N! in one line.
Sample Input
1
2
3
Sample Output
1
2
6
用斯特林公式$\begin{align}\large{}n! \approx \large{}\sqrt{2 \pi n}({\frac{n}{e}})^n\end{align}$估计一下阶乘位数,
其余没啥说的,测模板。。
#include<algorithm>
#include<iostream>
#include<istream>
#include<ostream>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<cstdio>
#include<string>
using std::max;
using std::cin;
using std::cout;
using std::endl;
using std::swap;
using std::string;
using std::istream;
using std::ostream;
struct BigN {
typedef unsigned long long ull;
static const int Max_N = ;
int len, data[Max_N];
BigN() { memset(data, , sizeof(data)), len = ; }
BigN(const int num) {
memset(data, , sizeof(data));
*this = num;
}
BigN(const char *num) {
memset(data, , sizeof(data));
*this = num;
}
void cls() { len = , memset(data, , sizeof(data)); }
BigN& clean(){ while (len > && !data[len - ]) len--; return *this; }
string str() const {
string res = "";
for (int i = len - ; ~i; i--) res += (char)(data[i] + '');
if (res == "") res = "";
res.reserve();
return res;
}
BigN operator = (const int num) {
int j = , i = num;
do data[j++] = i % ; while (i /= );
len = j;
return *this;
}
BigN operator = (const char *num) {
len = strlen(num);
for (int i = ; i < len; i++) data[i] = num[len - i - ] - '';
return *this;
}
BigN operator + (const BigN &x) const {
BigN res;
int n = max(len, x.len) + ;
for (int i = , g = ; i < n; i++) {
int c = data[i] + x.data[i] + g;
res.data[res.len++] = c % ;
g = c / ;
}
while (!res.data[res.len - ]) res.len--;
return res;
}
BigN operator * (const BigN &x) const {
BigN res;
int n = x.len;
res.len = n + len;
for (int i = ; i < len; i++) {
for (int j = , g = ; j < n; j++) {
res.data[i + j] += data[i] * x.data[j];
}
}
for (int i = ; i < res.len - ; i++) {
res.data[i + ] += res.data[i] / ;
res.data[i] %= ;
}
return res.clean();
}
BigN operator * (const int num) const {
BigN res;
res.len = len + ;
for (int i = , g = ; i < len; i++) res.data[i] *= num;
for (int i = ; i < res.len - ; i++) {
res.data[i + ] += res.data[i] / ;
res.data[i] %= ;
}
return res.clean();
}
BigN operator - (const BigN &x) const {
assert(x <= *this);
BigN res;
for (int i = , g = ; i < len; i++) {
int c = data[i] - g;
if (i < x.len) c -= x.data[i];
if (c >= ) g = ;
else g = , c += ;
res.data[res.len++] = c;
}
return res.clean();
}
BigN operator / (const BigN &x) const {
return *this;
}
BigN operator += (const BigN &x) { return *this = *this + x; }
BigN operator *= (const BigN &x) { return *this = *this * x; }
BigN operator -= (const BigN &x) { return *this = *this - x; }
BigN operator /= (const BigN &x) { return *this = *this / x; }
bool operator < (const BigN &x) const {
if (len != x.len) return len < x.len;
for (int i = len - ; ~i; i--) {
if (data[i] != x.data[i]) return data[i] < x.data[i];
}
return false;
}
bool operator >(const BigN &x) const { return x < *this; }
bool operator<=(const BigN &x) const { return !(x < *this); }
bool operator>=(const BigN &x) const { return !(*this < x); }
bool operator!=(const BigN &x) const { return x < *this || *this < x; }
bool operator==(const BigN &x) const { return !(x < *this) && !(x > *this); }
};
istream& operator >> (istream &in, BigN &x) {
string src;
in >> src;
x = src.c_str();
return in;
}
ostream& operator << (ostream &out, const BigN &x) {
out << x.str();
return out;
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
int n;
while (~scanf("%d", &n)) {
BigN res = ;
if ( == n || == n) {
puts("");
continue;
}
for (int i = ; i <= n; i++) res *= i;
cout << res << endl;
}
return ;
}
hdu 1042 N!的更多相关文章
- HDU 1042 N! 參考代码
HDU 1042 N! 题意:给定整数N(0 ≤ N ≤ 10000), 求 N! (题目链接) #include <iostream> using namespace std; //每一 ...
- HDU 1042 大数阶乘
B - 2 Time Limit:5000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Statu ...
- hdu 1042 N!(高精度乘法 + 缩进)
题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1042 题目大意:求n!, n 的上限是10000. 解题思路:高精度乘法 , 因为数据量比较大, 所以 ...
- N! HDU 1042
N! Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submiss ...
- Hdu 1042 N! (高精度数)
Problem Description Givenan integer N(0 ≤ N ≤ 10000), your task is to calculate N! Input OneN in one ...
- HDU 1042 N!(高精度计算阶乘)
N! Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submiss ...
- hdu 1042 N!(大数的阶乘)
N! Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submiss ...
- hdu 1042
貌似之前也写过这个题目的解题报告...老了,记性不好 从贴一遍吧! 代码理解很容易 AC代码: #include <iostream> #include <stdio.h> # ...
- hdu 1042 N!(高精度乘法)
Problem Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N! Input One N in ...
随机推荐
- Android基础总结(9)——网络技术
这里主要讲的是如何在手机端使用HTTP协议和服务器端进行网络交互,并对服务器返回的数据进行解析,这也是Android最常使用到的网络技术了. 1.WebView的用法 Android提供的WebVie ...
- extern “C”调用测试与验证-2016.01.06
1 调用情形说明 在上一篇关于extern “c”原理以及用法中,详细的说明了为什么需要extern “c”以及如何使用它解决c与c++混合编程时遇到的问题.接下来,使用示例验证方式验证c与c++函数 ...
- openstack实例热迁移
[DEFAULT]scheduler_default_filters=AllHostsFilterallow_resize_to_same_host=Trueallow_migrate_to_same ...
- 网络设备模拟器 GNS3
https://www.gns3.com/support/docs/linux-installation sudo dpkg --add-architecture i386 sudo add-apt- ...
- SVN创建资源与分支详解
创建分支的意义: 简单说,分支就是用于区分开发版本与当前发布版本的. 1. 主干负责新功能的开发 2..分支负责修正当前发布版本的bug(对于可以放入下个发布版本的改进性bug可以直接在主干上开发) ...
- 去HTML代码
Code: public static string NoHTML(string Htmlstring) { //删除脚本 Htmlstring = Regex.Replace(Htmlstring, ...
- 009Linux密码故障排除
1.Root密码破解/忘记Root密码: 步骤: (1)在系统启动时进入grub选项菜单: 在系统开机读秒时,按回车键,注意,要迅速,读秒的时间很快,但还需注意的是,虽然需要迅速,但是只按一次回车键就 ...
- 设备版本,设备号,APP版本,APP名称获取
//获取设备id号 UIDevice *device = [UIDevice currentDevice];//创建设备对象 NSString *deviceUID = [[NSString allo ...
- [原]Hrbust1328 相等的最小公倍数 (筛素数,素因子分解)
本文出自:http://blog.csdn.net/svitter/ 题意: 求解An 与 An-1是否相等. n分为两个情况-- 1.n为素数, 2.n为合数. = =好像说了个废话..素数的时候 ...
- 树莓派 B+ Yeelink实现图像监控
树莓派 B+ Yeelink实现图像监控 数值传感器请参考 : http://blog.csdn.net/xiabodan/article/details/39084877 1 安装摄像头 ...