Description(高精度的一類問題)

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 <= 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
下面提供了一種思路代碼如下：

include

using namespace std;

define SIZE 10000

int main()
{
char c[SIZE];
int n;
while(cin>>c>>n)
{
int i=0,j,arr[SIZE]={0},len=strlen(c),max=0,set=0,temp,k,record=0;//arr[]中所有的元素初始化為0
for(i=0;i
{
if(c[i]=='.')
record=len-i-1;
else
max=max10+c[i]-'0';
}
if(record)
{
for(k=len-1;c[k]=='0';k--,record--)
max/=10;//去掉后面多余的0
}
for(i=0,temp=max;i
arr[i]=temp;//將數(shù)倒置導(dǎo)入數(shù)組
record=recordn;//record為總的小數(shù)位
for(i=0;i
{
for(j=0;j
{
arr[j]=arr[j]*max+set;
set=arr[j]/10;
arr[j]=arr[j];
}
}
for(i=SIZE-1;arr[i]==0;i--)
if(arr[i]!=0)
break; //注意此時計算結(jié)果仍是倒置的玲昧，并且把小數(shù)點前無用的零去掉
if(record>i)
{
cout<<'.';
for(k=record-1;k>=0;k--)
cout<<arr[k];
}
else
{
for(;i>=0;i--)
{
cout<<arr[i];
if(i==record&&record!=0)
cout<<'.';
}
}
cout<<endl;
}
return 0;
}