poj1001

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字数统计: 678 | 阅读时长 ≈ 4

Title: 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

C++ Code

Memory: 184K Time: 0MS

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

int main(int argc, char const *argv[])
{
    string s;
    int n = 0;

    while (cin >> s >> n)
    {
        int end = s.length();
        int point = -1;
        //检测小数点
        for (int i = 0; i < end; i++)
            if (s[i] == '.')
            {
                point = i;
                break;
            }
        //去除小数点后的尾随0
        if (point != -1)
            while (end > point && s[end - 1] == '0')
                end--;
        //计算最终结果小数点的位置，point应为小数点后数字的数量
        if (point != -1)
            point = n * (end - point - 1);
        int *nums = new int[6];
        int *ans = new int[160];
        int *temp = new int[160];
        for (int i = 0; i < 6; i++)
            nums[i] = 0;
        for (int i = 0; i < 160; i++)
        {
            ans[i] = 0;
            temp[i] = 0;
        }
        int tl = 6;
        for (int i = end - 1, j = 0; i >= 0; i--, j++)
        {
            if (s[i] == '.')
            {
                j--;
            }
            else
                nums[j] = s[i] - '0';
        }
        for (int i = 0; i < 6; i++)
            ans[i] = nums[i];
        //计算
        int w = 0;
        int r = 0;
        for (int i = 1; i < n; i++)
        {
            tl = 6 * i;
            for (int m = 0; m < tl; m++)
            {
                temp[m] = ans[m];
                ans[m] = 0;
            }
            int j = 0, k = 0, watch = 0;
            for (j = 0; j < 6; j++)
            {
                for (k = 0; k < tl; k++)
                {
                    r = nums[j] * temp[k];
                    watch = w + r + ans[j + k];
                    ans[j + k] = watch % 10;
                    w = watch / 10;
                }
            }
            while (w != 0)
            {
                ans[j + k - 1] = w % 10;
                w = w / 10;
                k++;
            }
            for (j = 0; j < 160; j++)
                temp[j] = ans[j];
        }
        //将结果复制到string中
        bool flag = false;
        string answer;
        for (int i = 159; i >= 0; i--)
        {
            if (ans[i] != 0)
                flag = true;
            if (flag == true)
                answer.push_back('0' + ans[i]);
        }
        //输出0
        if (answer.length() == 0)
        {
            answer = "0";
            cout << answer << endl;
            continue;
        }
        //为位数不足的小数添加足够的0
        if (point > (int)answer.length())
        {
            answer.insert(0, point - answer.length(), '0');
        }
        //point不是-1和0的时候添加小数点
        if (point >= 1)
            answer.insert(answer.length() - point, 1, '.');

        cout << answer << endl;
        delete[] nums;
        delete[] ans;
        delete[] temp;
    }
    return 0;
}