A - 高精度（大数）N次方（第二季水）

Posted 2020-06-13

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 R ⁿ 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<stdio.h>
using namespace std;
int main()
{
    char a[10];
    int n;
    while(cin>>a>>n){
    int b=0,c[250]={0},i,j,x,flag=0;
    c[249]=1;
    for(i=0;a[i]!=‘\0‘;i++)
    {
        if(a[i]==‘.‘)
        {
            x=i;         //x为小数点所在的位置
            flag=1;
            continue;
        }
        b=b*10+a[i]-‘0‘;   //b为a[10]去掉小数点后的整型数字
    }
    x=i-x-1;              //x为小数部分的位数
    for(i=1;i<n+1;i++)
    {
       for(j=0;j<250;j++)
           c[j]*=b;
       for(j=249;j>=0;j--)
       if(c[j]>=10)
       {
           c[j-1]+=c[j]/10;
           c[j]%=10;
       }
    }
    if(flag==0)
    {
        for(i=0;c[i]==0;i++);       //i为c[250]中才首位开始判断 不是0的那一位
        for(;i<250;i++)
           cout<<c[i];              //无小数部分时直接输出c[250]，无需考虑小数点
    }
    else if(a[0]==‘0‘)             //只有小数部分时
    {
        flag=0;  
                    //整数部分为0，不输出
       for(j=i;j<250;j++)
            if(c[j]!=0)
            {
                flag=1;
                break;                //利用flag的值来判断结果是否存在小数部分
            }
        if(flag==1)
        {
            cout<<".";
            for(j=249;c[j]==0&&j>=i;j--);        //去掉小树部分最后多余的0
            for(i=250-n*x;i<=j;i++)
                cout<<c[i];
        }
    }
    else                     //既有小数部分也有整数部分
    {
        flag=0;
        for(i=0;c[i]==0;i++);
        for(;i<250-n*x;i++)        //小数部分一共是n*x位，之前全为整数部分
            cout<<c[i];
        for(j=i;j<250;j++)
            if(c[j]!=0)
            {
                flag=1;
                break;            //此处 同上述只有小数部分的情况
            }
        if(flag==1)
        {
            cout<<".";
            for(j=249;c[j]==0&&j>=i;j--);
            for(;i<=j;i++)
                cout<<c[i];
        }

    }
    cout<<endl;
    }
    system("pause");
    return 0;
}