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
- 【LeetCode】17. Letter Combinations of a Phone Number
题目: 思路:设置两个List,一个存储当前层,一个存储最终层 public class Solution { public List<String> letterCombinations ...
- c语言描述简单的线性表,获取元素,删除元素,
//定义线性表 #define MAXSIZE 20 typedef int ElemType; typedef struct { ElemType data[MAXSIZE]; //这是数组的长度, ...
- leetcode279. Perfect Squares
learn from DP class Solution { public: int numSquares(int n) { if(n<=0)return 0; int * dp = new i ...
- CentOS6.5 ssh远程连接缓慢解决方法
UseDNS no GSSAPIAuthentication no 1.适用命令及方案如下:[远程连接及执行命令]ssh -p22 root@10.0.0.19ssh -p22 root@10.0.0 ...
- 【MySQL】MySQL同步报错-> received end packet from server, apparent master shutdown: Slave I/O thread: Failed reading log event, reconnecting to retry报错解决和分析
[root@db-ft-db-48 ~]# tail -f /mysqlLog/beside_index_err.log 140102 20:42:26 [Note] Slave: received ...
- hadoop-集群管理(2)——内存设置
http://www.cnblogs.com/yuechaotian/archive/2013/03/08/2949607.html 1. 内存 hadoop为各个守护进程(namenode,seco ...
- c programming language ___ 5_2.c
#include <stdio.h> #include <ctype.h> #define BUG printf("here!bug!\n"); int g ...
- Android IOS WebRTC 音视频开发总结(三九)-- win10升级为何要p2p
本文主要介绍webrtc p2p的应用场景,文章来自博客园RTC.Blacker,支持原创,转载请说明出处. P2P最简单的解释就是两个客户端之间直接进行数据交互,不经过服务端转发. 最早接触P2P是 ...
- linux驱动程序框架基础
============================ 指引 ============================= 第一节是最基础的驱动程序: 第二节是/dev应用层接口的使 ...
- 模拟cpu调度
先来先服务实现简单但是平均周转时间过长 短作业优先算法缩短了平均周转时间 #!/usr/bin/python #-*- coding: utf-8 -*- # # table # 0:进程号 1:到达 ...