UVa 748 - Exponentiation

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本文介绍了一种用于计算大数幂的精确方法，并提供了一个具体的C语言程序示例。该程序能够处理非常大的基数和指数，确保计算结果的准确性。文章通过样例输入输出展示了如何解决实际问题。

Exponentiation

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 ofRⁿ 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

Miguel A. Revilla
2000-02-09

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

void multiply (char n[],char s[],char m[])
{
    int i=0,j=0,l,k,p1,p2,len1,len2,r[600];
    memset(r,0,sizeof(int)*600);
    len1=strlen(n);
    len2=strlen(s);
    while(n[i]!='.')i++;
    p1=len1-i-1;
    while(s[j]!='.')j++;
    p2=len2-j-1;
    for(i=len1-1,l=len1-1;i>=0;i--)
        if(n[i]!='.')
        for(j=len2-1,k=599-(len1-1-l--);j>=0;j--)
            if(s[j]!='.')
                r[k--]+=(n[i]-48)*(s[j]-48);
    for(i=599;i>=0;i--)
        if(r[i]>9)
        {
            r[i-1]+=r[i]/10;
            r[i]%=10;
        }
    k=0;
    memset(m,0,sizeof(char)*600);
    while(r[k]==0&&k<=598&&k<600-p1-p2)k++;
    m[600-k-p1-p2]='.';
    for(i=k,j=0;i<=599;j++)
            if(m[j]!='.')
                m[j]=r[i++]+48;
    k=strlen(m)-1;
    while(m[k]=='0')k--;
        m[++k]='\0';
}

int main()
{
    int n;
    char s[600],m[600];
    while(scanf("%s %d",s,&n)!=EOF)
        {
            memset(m,0,sizeof(char)*600);
            m[0]='1';
            m[1]='.';
            m[2]='0';
            while(n--)
            {
                multiply(m,s,m);
            }
    if(m[strlen(m)-1]=='.')
        m[strlen(m)-1]='\0';
           printf("%s\n",m);
        }
    return 0;
}