uva748 - Exponentiation 高精度小数的幂运算
| Exponentiation |
Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.
This problem requires that you write a program to compute the exact value of Rn where R is a real number (0.0 < R < 99.999) and n is an integer such that
.
Input
The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.
Output
The output will consist of one line for each line of input giving the exact value of Rn. Leading zeros and insignificant trailing zeros should be suppressed in the output.
Sample Input
95.123 12
0.4321 20
5.1234 15
6.7592 9
98.999 10
1.0100 12
Sample Output
548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201
套用模版,注意小数的位数不足时要补0,后续0要清掉(其实对底数进行后续0清除就好了,因为出现后续0的唯一可能就是底数有后续0)
/*
高精度的幂。幂为低精度。
*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <climits>
using namespace std; #define maxn 30000 struct bign
{
int len, s[maxn]; bign()
{
memset(s, , sizeof(s));
len = ;
} bign(int num)
{
*this = num;
} bign(const char* num)
{
*this = num;
} bign operator = (int num)
{
char s[maxn];
sprintf(s, "%d", num);
*this = s;
return *this;
} bign operator = (const char* num)
{
len = strlen(num);
for (int i = ; i < len; i++) s[i] = num[len-i-] - '';
return *this;
} string str() const
{
string res = "";
for (int i = ; i < len; i++) res = (char)(s[i] + '') + res;
if (res == "") res = "";
return res;
} bign operator + (const bign& b) const
{
bign c;
c.len = ;
for (int i = , g = ; g || i < max(len, b.len); i++)
{
int x = g;
if (i < len) x += s[i];
if (i < b.len) x += b.s[i];
c.s[c.len++] = x % ;
g = x / ;
}
return c;
} void clean()
{
while(len > && !s[len-]) len--;
} bign operator * (const bign& b)
{
bign c; c.len = len + b.len;
for (int i = ; i < len; i++)
for (int j = ; j < b.len; j++)
c.s[i+j] += s[i] * b.s[j];
for (int i = ; i < c.len-; i++)
{
c.s[i+] += c.s[i] / ;
c.s[i] %= ;
}
c.clean();
return c;
} bign operator - (const bign& b)
{
bign c; c.len = ;
for (int i = , g = ; i < len; i++)
{
int x = s[i] - g;
if (i < b.len) x -= b.s[i];
if (x >= )
g = ;
else
{
g = ;
x += ;
}
c.s[c.len++] = x;
}
c.clean();
return c;
} bool operator < (const bign& b) const
{
if (len != b.len) return len < b.len;
for (int i = len-; i >= ; i--)
if (s[i] != b.s[i]) return s[i] < b.s[i];
return false;
} bool operator > (const bign& b) const
{
return b < *this;
} bool operator <= (const bign& b)
{
return !(b > *this);
} bool operator == (const bign& b)
{
return !(b < *this) && !(*this < b);
} bool operator != (const bign& b)
{
return (b < *this) || (*this < b);
} bign operator += (const bign& b)
{
*this = *this + b;
return *this;
}
}; istream& operator >> (istream &in, bign& x)
{
string s;
in >> s;
x = s.c_str();
return in;
} ostream& operator << (ostream &out, const bign& x)
{
out << x.str();
return out;
} int main()
{
bign a,ans;
int b;
string c; while (cin >> c >> b)
{
ans = ; int index = c.find('.'); if (index != -)
{
//后续0
for (int i = c.size()-; i > index; --i)
{
if (c[i] == '')
c.erase(i, );
else
break;
} c.erase(index, );
} index = c.size()-index; index *= b; a = c.c_str(); a.clean(); for (int i = ; i < b; ++i)
{
ans = ans*a;
} string s = ans.str(); //补0
int dif = s.size()-index;
if (dif >= )
{
s.insert(s.end()-index, '.');
}
else
{
for (int i = ; i < -dif; ++i)
{
s = '' + s;
}
s = '.' + s;
} cout << s << endl;
}
}
uva748 - Exponentiation 高精度小数的幂运算的更多相关文章
- 【POJ 1001】Exponentiation (高精度乘法+快速幂)
BUPT2017 wintertraining(15) #6A 题意 求\(R^n\) ( 0.0 < R < 99.999 )(0 < n <= 25) 题解 将R用字符串读 ...
- hdu 1063 Exponentiation (高精度小数乘法)
//大数继续,额,要吐了. Problem Description Problems involving the computation of exact values of very large m ...
- BigDecimal类(高精度小数)
位置:java.math.BigDecimal 作用:提供高精度小数数据类型及相关操作 一.基本介绍 BigDecimal为不可变的.任意精度的有符号十进制数,其值为(unscaledValue * ...
- 算数运算符: + - * / //(地板除) %(取余) **(幂运算) / 比较运算符 > < >= <= == !=
# ### python运算符 #(1) 算数运算符: + - * / //(地板除) %(取余) **(幂运算) var1 = 5 var2 = 8 # +res = var1 + var2 pri ...
- Modular_exponentiation模幂运算
https://en.wikipedia.org/wiki/Modular_exponentiation 蒙哥马利(Montgomery)幂模运算是快速计算a^b%k的一种算法,是RSA加密算法的核心 ...
- POJ1026 Cipher(置换的幂运算)
链接:http://poj.org/problem?id=1026 Cipher Time Limit: 1000MS Memory Limit: 10000K Total Submissions ...
- 程序设计入门——C语言 第5周编程练习 1高精度小数(10分)
1 高精度小数(10分) 题目内容: 由于计算机内部表达方式的限制,浮点运算都有精度问题,为了得到高精度的计算结果,就需要自己设计实现方法. (0,1)之间的任何浮点数都可以表达为两个正整数的商,为了 ...
- 组合数学 - 置换群的幂运算 --- poj CARDS (洗牌机)
CARDS Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 1448 Accepted: 773 Description ...
- 迭代加深搜索 codevs 2541 幂运算
codevs 2541 幂运算 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 钻石 Diamond 题目描述 Description 从m开始,我们只需要6次运算就可以计算出 ...
随机推荐
- Twelves Monkeys (multiset解法 141 - ZOJ Monthly, July 2015 - H)
Twelves Monkeys Time Limit: 5 Seconds Memory Limit: 32768 KB James Cole is a convicted criminal ...
- Java中Object转化为int类型
转自:http://blog.sina.com.cn/s/blog_5f8421fb010162kb.html Java中由Object类型转化为int类型时,不能直接转化,先是将Object类型转化 ...
- 05-spring-bean注入
spring中只有两大核心技术: 控制反转(IOC)&依赖注入(DI),AOP(面向切面编程) 依赖注入:指利用配置文件的关系,来决定类之间的引用关系,以及数据的设置操作. 构造方法注入 默认 ...
- php fpm安装curl后,nginx出现connect() to unix:/var/run/php5-fpm.sock failed (13: Permission denied)的错误
这里选择直接apt-get安装,因为比起自己编译简单多了,不需要自己配置什么 #sudo apt-get install curl libcurl3 libcurl3-dev php5-curl 安装 ...
- 解决在IE9,IE10浏览器下,程序没有任何错误,easy ui页面不加载任何数据的问题
对于web应用程序,经常用到开发人员工具,按F12,可以调试脚本,可以查看监视网络,查看各页面加载时间,非常方便,今天在调试js时,不小心打开了兼容性视图, 之后每次打打开页面时,均不显示页面post ...
- SQL语句创建相同结构的表
--Oracle的语句create table sa_salaryRecord as select * from sa_salary where 1=2; --MSSQL的语句select * int ...
- linux下Samba服务配置
SMB是基于客户机/服务器型的协议,因而一台Samba服务器既可以充当文件共享服务器,也可以充当一个Samba的客户端,例如,一台在Linux 下已经架设好的Samba服务器,windows客户端就可 ...
- error while loading shared libraries错误解决
在编译引用了第三方库的代码后,执行出现了以下错误 [work@xxx zktest]$ ./a.out ./a.out: error while loading shared libraries: l ...
- Qt 槽函数的使用
今天在代码中遇到这样一个问题,自己感觉槽和函数都写的没错,但是就是不执行槽函数,因为是一个定时器的使用,即定时时间到了就执行槽函数. SeventhWizardPage::SeventhWizardP ...
- ITDB部署
官方地址:http://www.sivann.gr/software/itdb/ 方法如下: 前提:首先需要三个东西:APACHE,PHP5,SQLITE3,php5-sqlite 环境:ubuntu ...