hdu 1063 Exponentiation (高精度小数乘法)

//大数继续，额，要吐了。

Problem 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 <= 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

 

Source

East Central North America 1988

 

/********************

高精度数，模拟乘法，

*********************/

Code:

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
#define N 1000
struct Bint
{
    int num[N]; //数值存储
    int p;//整数位数
    int del;//小数位数
    int len;// 总的数位长度
};
int Find(char a,char str[])//  找小数点的位置,实际返回的是整数部分的位数
{
    int i,len = strlen(str);
    for(i = 0;i<len;i++)
        if(str[i]==a)
            return i;
    return len;//  整数时直接返回  长度
}

void print(struct Bint s)//  输出函数
{
    int i;
    /*
    if(s.p=1&&s.num[s.len-1]==0)
        s.p-=1;
        */
    for(i = s.p+s.del-1;i>=s.del;i--)
        printf("%d",s.num[i]);
    if(s.del>0){
        printf(".");
        for(i = s.del-1;i>=0;i--)
            printf("%d",s.num[i]);
    }
    printf("
");
}
void trans(Bint &a,char str[])//  将输入的字符串形式的数 转换为结构体
{
    memset(a.num,0,sizeof(a.num));
    int len = strlen(str);
    a.p = Find('.',str);
    a.del = (len==a.p)?0:(len-a.p-1);//  如果包含小数点则小数位数为 数的长度-整数位数-1
    int i=0,k = 0,pa= a.p,l = len-1;
    while(str[i++]=='0'&&i<=pa)//   去除前导0 小数点前面的 0 也不需要保存
        a.p--;
    while(str[l--]=='0'&&l>=pa)//  去除 后导 0
        a.del--;
    for(i = pa+a.del;i>pa;i--)//  小数位 赋值
        a.num[k++] = str[i]-'0';
    for(i = pa-1;i>=pa-a.p;i--)
        a.num[k++] = str[i]-'0';
    a.len = k;
}
Bint  mul(Bint a,Bint b)// 模拟乘法，值存到结构体中
{
    Bint m;
    memset(m.num,0,sizeof(m.num));
    int i,j;
    for(i = 0;i<a.len;i++)
    {
        for(j = 0;j<b.len;j++)
        {
            m.num[i+j]+=a.num[i]*b.num[j];
        }
    }
    m.del = a.del+b.del;
    m.len = a.len+b.len-1;
    for(i = 0;i<m.len;i++)
        if(m.num[i]>=10)
        {
            m.num[i+1] += m.num[i]/10;
            m.num[i]%=10;
        }
    if(m.num[m.len]>0&&m.num[m.len]<10)
        m.len+=1;
    else if(m.num[m.len]>10)
    {
        m.num[m.len+1]+=m.num[m.len]/10;
        m.num[m.len]%=10;
        m.len+=2;
    }
    m.p = m.len-m.del;
    return m;
}
int main()
{
    char str[N];
    Bint a,s;
    int n,i;
    while(scanf("%s%d",str,&n)!=EOF)
    {
        trans(a,str);
        trans(s,str);
        for(i = 1;i<n;i++)
            s = mul(s,a);
        print(s);

    }
    return 0;
}