大数算法——高精度大数幂次

大数——高精度大数求幂

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

这道题有不少细节点需要注意的，自己代码写的很繁琐，没那么简化，应当适当优化才对。

1.注意第二个样例数据，当前面的整数部分为0时，不输出，

2.确保当输入数据为0时依旧有结果。

3.输入10和10.00验证答案是否达到题目要求，这也是坑

4.判断小数点位置时的算法要好好考虑，不是简单的从前往后数第几位是小数点，

代码如下：

#include <iostream>

#include <cstdio>

#include <string.h>

#include <string>

using namespace std;

void fun(char str1[] ,int n)//n为幂次,str1已经倒置所有数字

{

        int n1 = strlen(str1);

        int a[5000], b[5000],temp[5000];

        for (int i = 0; i < 5000; i++)

               a[i] = b[i] = 1000, temp[i] = 0;

        for (int i = 0; i < n1; i++)//char to int

        {

               a[i] = b[i] = str1[i] - '0';

        }

        int count = n1, k;

        for (int i = 1; i < n; i++)//幂次

        {

               

               for (int j = 0; j < n1; j++)//底下的数字

               {

                       k = j;

                       for (int j2 = 0; a[j2]!=1000; ++j2)//上面的数字

                       {

                              temp[k] += a[j2] * b[j];

                              k++;

                       }

               }

               for (int i =0; i<k; i++)

               {

                       temp[i + 1] += temp[i] / 10;

                       temp[i] = temp[i] % 10;

               }

               for (int i = 0; i < k || temp[i] != 0; i++)

               {

                       a[i] = temp[i];

                       temp[i] = 0;

               }

        }

        int stri = 0;

        for (int i = k + 1; i >= 0; i--)

        {

               if (a[i] != 1000)

               {

                       str1[stri] = a[i] + '0'; stri++;

               }

        }

}

int main()

{

        char s[5000]; int n;

        memset(s, '\0', sizeof(s));

        while (cin>>s>>n)

        {

               char str1[5000];

               memset(str1, '\0', sizeof(str1));     

               int k=0,pointset=0;

               for (int i = strlen(s)-1; i>=0; i--)//转换到char 字符数组

               {

                       if (s[i] != '.')

                       {

                              str1[k] = s[i];

                              k++;

                       }

                       else pointset = strlen(s)-1-i;

               }

               fun(str1, n);

               pointset *= n;//小数点位置

               int t1 = strlen(str1),zero=t1;

               if(strchr(s,'.')!=NULL)

               for (int i = t1 - 1; i >= t1-pointset; i--)

               {

                       if (str1[i] != '0')break;

                       else zero = i;//最后尾数的0的清除

               }

               int flag = 0;

               if(pointset)

               for (int i = 0; i <= t1-pointset; i++)

               {

                       if (str1[i] == '0')

                              flag = i;//开头的0的个数

                       else break;

               }

               for (int i = flag; i < t1; i++)

               {

                       if (zero == 0) { cout << 0; break; }

                       if (zero == i ) { break; }

                              if (i == t1 - pointset&&pointset)

                              {

                                      cout << ".";

                              }

                              cout << str1[i];

               }

               cout << endl;

        }

        return 0;

}