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.

可以直接运行
下面是输入和输出
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
Hint

#include<stdio.h>
#include<string.h>

   
char *multiply1(char *numberN,char *numberM);
void mi();
int main()
{
/*    char numberN[1500], numberM[1500];//为什么这里不能定义为char *num呢？
    scanf("%s%s", numberN, numberM);
    char *j;
    j=multiply1(numberN,numberM);
    puts(j);
*/
while(1)
 mi();
 
}
void mi()
{
    char di[1500];
    int mici;
    printf("输入底数，0.0-99.999；输入幂，1-25\n ");
    scanf("%s",di);
    scanf("%d",&mici);
    int i;
    char *qu[mici];//定义一个char 数组，用于存储不同幂次下的值
    qu[0]=di;

    for(i=0;i<mici-1;i++)
    {
        
   
    qu[i+1]=multiply1(qu[i],di);
    }
    puts(qu[mici-1]);
}
char *multiply1(char *numberN,char *numberM)//高精度乘法
{
    int n1,n2;//n1,n2是小数点位置。
    int i, j;
    char w[1500];
    int k=0;
    int n = strlen(numberN), m = strlen(numberM);//求字符串长度

   
   int a[n]={0}, b[m]={0};//这样就先把数组置为全0
 
    for (i = 0, j = n - 1; j>=0; j--) {
        if(numberN[j]!='.')//要找出小数点的位置
        {
           a[i] = numberN[j] - '0';
           i++;
        }
        else
        n1=i;
    }

   for (i = 0, j = m - 1;j>=0;j--) {
        if(numberM[j]!='.')
        {
           b[i] = numberM[j] - '0';
           i++;
         }
         else
         n2=i;
    }
    int c[3000];
    for (i = 0; i < 3000; i++) {
        c[i] = 0;
    }
  
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            c[i + j] += a[i] * b[j];
        }
    }   
    for (i = 0; i < n + m-1; i++) {//这里的总位数是否是n+m-1
        if (c[i] >= 10) {
            c[i + 1] += c[i] / 10;
            c[i] %= 10;
        }
    }
     

        for (j = 2999; j > 0; j--) {
        if (c[j] != 0)
        break;
    }

if(n1+n2>j)
{
    w[0]='0';
    w[1]='.';
    k=2;
    for(i=1;i<=n1+n2-j-1;i++)
  {
     w[k]='0';
     k++;
  }
  for(i=j;i>=0;i--)
  {
      w[k]=48+c[i];
      k++;
  }
}
else
{

   i=j;

  while(i>=0)
  {
    if(i!=n1+n2-1)
    {
        w[k]=48+c[i];
        i--;
        k++;
    }
    else
    {
      w[k]='.';
      k++;
      w[k]=c[i]+48;
      i--;
    }
 }
}
return w;
}

我这程序可以运行，只是数据结果和题目给出的结果有偏差，位数相同。
那个乘法函数是先转化为整数的高精度程度，然后找出小数点的位数，然后插入小数点，最后返回的仍然是两个含小数的两个数相乘得到的结果，用字符数组返回。
由于幂次会大于2次，就反复调用这个乘法函数