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 $0 < n \le 25$.

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

95.123 120.4321 205.1234 156.7592  998.999 101.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721.0000000514855464107695612199451127676715483848176020072635120383542976301346240143992025569.92857370126648804114665499331870370751166629547672049395302429448126.76412102161816443020690903717327667290429072743629540498.1075960194566517745610440100011.126825030131969720661201

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;}