【原】 POJ 1001 Exponentiation 大整数乘法 解题报告

http://poj.org/problem?id=1001

方法：

大整数乘法

这题没什么好说的，直接用字符串模拟整个过程

 

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

#include <iostream>
#include <string>

using namespace std ;

//大浮点数相乘，此时系统提供的精度远远不能达到要求

string multiply_bignum(const string& lhs,const string& rhs)
{
    //if( lhs=="0" || rhs=="0" )
    //    return "0" ;

    //lengths of the two operators , including dot
    size_t len1 = lhs.size() ;
    size_t len2 = rhs.size() ;
    int index=0 ;

    //carefully judge whether exist 0
    //.000 or 0.000 or 000.
    for( index=0 ; index<len1 ; ++index)
    {
        if( lhs[index]!='0' && lhs[index]!='.' )
            break;
    }
    if( index==len1 )
        return "0" ;

    for( index=0 ; index<len2 ; ++index)
    {
        if( rhs[index]!='0' && rhs[index]!='.' )
            break;
    }
    if( index==len2 )
        return "0" ;

    //copy the two operators , because later may delete their dots
    string s1(lhs);
    string s2(rhs);
    //max length of the result
    size_t lenResult = len1+len2 ;

    int val1,val2,tmpval,mul_carry,sum_carry ;
    int i,j,k,t,m ;

    //initialize the result to 0
    string result(lenResult,'0') ;

    size_t dotnum1,dotnum2,dotnumResult;
    //find the positions of the dots in the two operators
    string::size_type pos1 = s1.find(".") ;
    string::size_type pos2 = s2.find(".") ;
    string::size_type dotpos ;

    if( pos1 == string::npos )   //there is no dot
        dotnum1 = 0 ;
    else                         //the operator is float
    {
        s1.erase( pos1 , 1 ) ;   //delete the dot , 12.34-->1234
        dotnum1 = len1-pos1-1 ;  //number of the digit behind the dot
        --len1 ;                 //now the length decrease 1
    }

    if( pos2 == string::npos )
        dotnum2 = 0 ;
    else
    {
        s2.erase( pos2 , 1 ) ;
        dotnum2 = len2-pos2-1 ;
        --len2 ;
    }

    dotnumResult = dotnum1+dotnum2 ;    //number of digit behind the dot in the result
    dotpos = lenResult-dotnumResult-1 ; //the position of the dot in the result

    for( i=len2-1 , k=lenResult-1 ; i>=0 ; --i ,--k )  //s1*s2
    {
        string tmpresult(lenResult,'0') ;
        mul_carry = 0 ;    //carry of multiply in calculating the tmpresult
        sum_carry = 0 ;    //carry of sum in calculating the final result
        val2 = s2[i]-'0';  //get the digit of s2 in reverse order to simulate the process of multiply

        for( j=len1-1 , t=k ; j>=0 ; --j , --t )
        {
            val1 = s1[j]-'0' ;
            tmpval = val1*val2 + mul_carry ;
            mul_carry = tmpval/10 ;
            tmpresult[t] = tmpval - 10*mul_carry + '0' ;
        }
        tmpresult[t] = mul_carry + '0' ;  //put the final carry to the tmpresult

        for( m=lenResult-1 ; m>=0 ; --m ) //update the result
        {
            tmpval = (result[m]-'0') + (tmpresult[m]-'0') + sum_carry ;
            sum_carry = tmpval/10 ;
            result[m] = tmpval - 10*sum_carry + '0' ;
        }
    }
    string::size_type resultBeginPos ;
    string::size_type resultEndPos ;
    if( dotnumResult == 0 )  //there is no dot
    {
        resultBeginPos = result.find_first_not_of('0') ;
        return result.substr(resultBeginPos); ;
    }
    else
    {
        result.insert( ++dotpos , 1 , '.' );  //insert the dot into result

        resultBeginPos = result.find_first_not_of('0') ;
        resultEndPos = result.find_last_not_of('0') ;
        if( resultBeginPos == dotpos )  //0.123000
            return result.substr( resultBeginPos , resultEndPos-resultBeginPos+1 ) ;
        else if( resultBeginPos < dotpos ) //12.3400 or 12340.0000
        {
            if( resultEndPos > dotpos )  //12.3400
                return result.substr( resultBeginPos , resultEndPos-resultBeginPos+1 ) ;
            if( resultEndPos == dotpos )  //12340.0000
                return result.substr( resultBeginPos , resultEndPos-resultBeginPos );
        }
    }
}

//divide and conquer
string power_bignum(const string& R,int n)
{
    if(n==0)
        return "1" ;
    if(R=="0")
        return "0" ;

    if(n==1)
        return multiply_bignum( R , "1" ) ;
    string tmpresult = power_bignum(R,n/2);
    if( n%2 != 0 )
        return multiply_bignum( R , multiply_bignum(tmpresult,tmpresult) );
    else
        return multiply_bignum(tmpresult,tmpresult) ;
}

void run1001()
{
    string r;
    int n;
    while(cin>>r>>n)
        cout<<power_bignum(r,n)<<endl;
}