Problem F: Exponentiation

最新推荐文章于 2025-03-23 09:36:43 发布

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原文链接：http://www.cnblogs.com/NYNU-ACM/p/4236896.html

本文详细介绍了一种解决大数幂运算的算法实现，通过去除小数点、整数部分相乘并再次添加小数点的方式，精确计算R^n（0.0<R<99.999且0<n≤25）的值。该算法适用于需要高精度计算的场景，如国家债务等经济计算。

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Problem F: Exponentiation
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 4 Solved: 2
[Submit][Status][Web Board] [Edit] [TestData]
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 $0 < n \le 25$.

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 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
主要思路是：几的几次方。实际上也就是大数相乘，先把小数点去掉，然后相乘，最后再把小数点加上。比如说的 a 的 n 次方，那先把字符数组a里面的东西，赋值给字符数组b，然后 a 和 b 进行运算，乘的结果放到sum数组里，然后整型的 sum 数组里的每个元素转换成字符型的，赋值给字符数组b，然后再和 a数组运算，运算的次数和你输入的几次方有关系。。然后再进行细节方面的问题……

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    char a[1000];
    char b[1000];//和字符数组a一样的
    int n,i,j,k,lena,lenb,g,g1,h,ji1,ji2;
    int sum[1001];
    char s[1002];
    while(scanf("%s%d",&a,&n)!=EOF)
    {
        if(n==1)
             cout<<a<<endl;
       else
       {
           int count=0;
        memset(sum,0,sizeof(sum));
        lena=strlen(a);
        for(i=0;i<lena;i++)
        {
           if(a[i]=='.')
           {
               count=lena-i-1;//小数点后面几位
            for(j=i;j<lena-1;j++)
            {
                a[j]=a[j+1];
            }
            a[j]='\0';
            lena--;          //变成没有小数点的数
            break;
           }
        }
        count=count*n;//经过处理小数点最终有count位
        for(i=0;i<lena;i++)
        {
            b[i]=a[i];
        }
        b[i]='\0';
        lenb=strlen(b);
         for(i=1;i<=n-1;i++)
           {
                memset(sum,0,sizeof(sum));
               g=1000;g1=1000;
          for(j=lenb-1;j>=0;j--)//下面的乘数
          {
              for(k=lena-1;k>=0;k--)//上面的乘数
              {
                  sum[g--]+=(a[k]-'0')*(b[j]-'0');
              }
              g1--;
              g=g1;
          }
          for(k=1000;k>=1;k--)
          {
              sum[k-1]+=sum[k]/10;
              sum[k]=sum[k]-sum[k]/10*10;
          }
         int start=0;  
         int l=0;
         while(!sum[start] &&start<=999)
             start++;
           for(h=start;h<=1000;h++)
           {
                 b[l++]=sum[h]+'0';
           }
           b[l]='\0';
          lenb=strlen(b);
      }
    if(count!=0)
    {
        for(g=0;g<=1000;g++)
            s[g]=sum[g]+'0';
        s[1002]='\0';
        for(i=1000;i>1000-count;i--)
            s[i+1]=s[i];
        s[i+1]='.';
      for(i=0;i<=1002;i++)
      {
          if(s[i]!='0')
          {
              ji1=i;
              break;
          }
      }
      for(i=1001;i>=0;i--)
      {
          if(s[i]!='0')
          {
              ji2=i;
              break;
          }
      }
      for(i=ji1;i<=ji2;i++)
          cout<<s[i];
      cout<<endl;
    }
    else
        cout<<b<<endl;
    }
   }
    return 0;
}