Exponentiation

最新推荐文章于 2022-07-29 20:54:31 发布

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最新推荐文章于 2022-07-29 20:54:31 发布

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分类专栏： ACM 文章标签： poj Exponentia

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本文介绍了一种使用字符串处理实现精确大数幂运算的方法，适用于计算非常大的实数R的n次幂(R^n)，其中R介于0.0到99.999之间，n为不超过25的正整数。通过逐位相乘的方式，避免了浮点数运算中可能出现的精度损失问题。

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题目链接：
POJ 1001 Exponentiation
Description
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 0 < n <= 25.

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 R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don’t print the decimal point if the result is an integer.

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

#include<iostream>
#include<string>
using namespace std;

void mul(string &a, string &b)
{
    int i, j, k, c;
    int n1 = a.size(), n2 = b.size();
    string res;
    char aa[300];//一开始是直接赋给A的。可是有莫名其妙的错误。后来就开个char数组。
    res.resize(n1 + n2);
    for (i = 0; i<res.size(); i++)
        res[i] = 0;
    for (i = 0; i<n1; i++)
    {
        for (j = k = 0; j<n2; j++)//高精度X
        {
            c = (a[n1 - i - 1] - '0')*(b[n2 - j - 1] - '0') + k;

            res[i + j] += c;

            k = res[i + j] / 10;
            res[i + j] %= 10;
        }
        if (k)
            res[i + j] = k;

    }
    for (i = res.size() - 1; i>0 && res[i] == 0; i--)//清除0；
        res.erase(i, 1);
    for (i = 1; i <= res.length(); i++)
    {
        aa[i - 1] = (char)(res[res.length() - i] + '0');//反转，存储
    }

    aa[i - 1] = 0;

    a = string(aa);//赋值

}


int main()
{
    int n, i, j, pos;
    string in;


    while (cin >> in >> n)
    {
        for (pos = 0, i = 0; i<in.size(); i++)
        if (in[i] == '.')break;

        pos = in.size() - i - 1;//记录小数点位置。

        in.erase(i, 1);//在位置i删除长度为1的字符，就是删除小数点。

        string res("1"); 
        for (i = 0; i < n; i++) mul(res, in);

        if (pos != -1)pos = n * pos;

        while (res.size() < pos)

            res.insert(res.begin(), '0');//若长度小于精度，在前面添0

        if (pos != -1)
        {
            res.insert(res.end() - pos, '.');//插入小数点
            for (i = res.length() - 1; res[i] == '0'; i--);//除去多余的0；
            res.erase(i + 1);
            if (res[i] == '.')res.erase(i);//若是整数，除去小数点。
        }
        cout << res << endl;
    }
    return 0;
}