Exponentiation

最新推荐文章于 2022-07-29 20:54:31 发布

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本文介绍了一种算法，用于计算任意小数与整数之间的幂次运算，并确保输出结果的精确性。通过输入一对数值（小数与整数），程序能够准确地计算出结果，并在输出时去除不必要的零，保持结果简洁明了。

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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

Sample Output

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

HINT

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
	char str[1000],a[1000],a1[1000];
	int c;
	int n,i,j,k,t,x,num,t1,t2,y,y1,y2;
	while(cin>>str>>n)
	{
	
		x=0;num=0;
		t=strlen(str);
		for(i=0;i<t;i++)
		{
			if(str[i]=='.')
			{
				num=t-1-i;
				i++;
			}
			a[x]=str[i];
			a1[x++]=str[i];
		}
		a[x]='\0';
		a1[x]='\0';
		num*=n;
		int x1=x;
		for(i=0;i<n-1;i++)
		{	
			int sum[1000]={0};
			t1=999;t2=999;
			for(j=x-1;j>=0;j--)
			{
				for(k=x1-1;k>=0;k--)
					sum[t1--]+=((a[j]-'0')*(a1[k]-'0'));
				t2--;
			    t1=t2;
			}
		    c=0;
			for(y=999;y>=0;y--)
			{
				sum[y]+=c;
				c=0;
				if(sum[y]>9)
				{
					c=sum[y]/10;
					sum[y]=sum[y]%10;
				}
			}
			for(y1=0;y1<1000;y1++)
			{
				if(sum[y1]!=0)
					break;
			}
			x1=0;
			for(y2=y1;y2<1000;y2++)
				a1[x1++]=sum[y2]+'0';
			a1[x1]='\0';
		}
		if(num==0)
			cout<<a1<<endl;
		else
		{
			t=strlen(a1);
			if(a1[t-1]=='0')//若(2.0, 2)则输出4而不是4.00，多余的零去掉
			{
				for(i=t-1;i>=0;i--)
				{
					if(a1[i]!='0')
						break;
					num--;//小数点的位数
				}
				t=i+1;
			}
			for(i=0;i<t-num;i++)
				cout<<a1[i];//输出小数点之前的
			cout<<".";
			while(num>t-i)//若整数部分为0,小数部分不够填补小数点后的位数0
			{
				cout<<"0";
				num--;
			}
			for(j=i;j<t;j++)
				cout<<a1[j];//输出小数点后的
			cout<<endl;
		}
	}
	return 0;
}