Problem F: Exponentiation大数求幂

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
思路：注意当输入1.10000时我是先变成1.1处理

#include<stdio.h>
#include<iostream>
using namespace std;
#include<string.h>
void chengfa(char a[],char b[])
{
    int sum[1001];
    int lena=strlen(a)-1;
    int lenb=strlen(b)-1;
    int t1=lena;
    int t;
    memset(sum ,0,sizeof(sum));
    for(int i=lena;i>=0;i--)
        for(int j=lenb,t=1000-(t1-i);j>=0;j--)
            sum[t--]+=(a[i]-'0')*(b[j]-'0');
    for(int k=1000;k>=1;k--)
     {
        sum[k-1]+=sum[k]/10;
        sum[k]=sum[k]%10;
     }
     int start=0;
     while(start<=1000&&!sum[start])
        start++;
        memset(a,0,sizeof(a));
        int t2=0;
     for(int k=start;k<=1000;k++)
        a[t2++]=sum[k]+'0';
        a[t2]='\0';
 
}
int main()
{
    char a[1001],b[1001];
    char a1[1001],a2[1003];
    int n;
    memset(b,0,sizeof(b));
    while(cin>>a>>n)
    {
        int k6;
        int len6=strlen(a);
        for(k6=len6-1;k6>=0;k6--)
            if(a[k6]!='0')
           {
            a[k6+1]='\0';
            break;
           }
        int len1=strlen(a);
        int k;
        for(int i=0;i<len1;i++)
            if(a[i]=='.')
        {    k=(len1-1)-i;
            int j;
            for(j=i;j<len1;j++)
                a[j]=a[j+1];
                a[j]='\0';
                 len1--;
                 break;
 
        }
          int num=k*n;
        strcpy(b,a);
            n=n-1;
            while(n--)
           chengfa(a,b);
           int len2=strlen(a);
           int count1=0;
         int k1,k2,k3,k4;
           int lena=strlen(a)-count1;
        if(num<lena)                           没有前导0，即第一个是数字是大于0
           {
               for(k1=0;k1<lena-num;k1++)
                cout<<a[k1];
                 cout<<'.';
                 for(k2=k1;k2<lena;k2++)
                    cout<<a[k2];
                    cout<<endl;
           }
           else                                //有前导0
           {
 
               printf(".");
               for(k3=num-lena;k3>0;k3--)
                    printf("0");
               cout<<a;
               cout<<endl;
           }
    }
    return 0;
}