UVA 787 and Big Number Multiplication

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UVA 787 and Big Number Multiplication

Problem



Background: Big Numbers in this problem

Big numbers are effectively storing sequence of digit chunks and directly manipulating them as if they were a large number. The most intuitive way to do this is to make an array where every element is a digit, but this is not effectively using the memory. Hence, it is logical to store numbers "millions at a time", i.e.

324876498726832746
 [2]  | [1] | [0]

like that.

Necessities to Implement

Notice that although this problem may LOOK like DP, it's really solvable by O(N^2) enumeration. Also notice that in this enumeration we will only be multiplying a large number with a normal number. Pseudocode for algorithm:

CREATE LargeNumber NAME best SET -INFINITY
FOR EACH i BETWEEN 0 and LENGTH OF list DO:
    CREATE LargeNumber NAME product SET 1
    FOR EACH j BETWEEN i AND LENGTH OF list DO
        SET product TO product TIMES list[j]
        IF product IS GREATER THAN best DO
           SET best TO product

From here it's obvious the only methods we need to implement is initialization, equation, multiplication and greater-comparison.

My code demonstrates a simple, minimalist, object-oriented implementation.

Code

#include <iostream>
#include <algorithm>
#include <cstring>
#include <iomanip>
using namespace std;

long long arr[105];
long long preProd[105];

#define LEN 500
#define MOD 1000000
#define MS 6

struct Large
{
int64_t digits[LEN];
int sign; // -1, 0, or 1
Large()
{
memset(digits, 0, sizeof digits);
sign = 0;
}
Large& operator=(Large& other)
{
if(this != &other)
{
sign = other.sign;
memcpy(digits, other.digits, sizeof digits);
}
return *this;
}
Large& operator=(int other)
{
memset(digits, 0, sizeof digits);
if(other == 0) sign = 0;
else if(other > 0) sign = 1;
else sign = -1;
other = sign*other; // now it's positive
digits[0] = other%MOD;
digits[1] = other/MOD;
digits[2] = digits[1]/MOD;
digits[1] %= MOD;
return *this;
}
void operator*=(int other)
{
int osign;
if(other == 0) osign = 0;
else if(other > 0) osign = 1;
else osign = -1;
sign = sign*osign;
for(int i=0; i<LEN; i++)
digits[i] = digits[i]*osign*other;
for(int i=0; i<LEN-1; i++)
{
digits[i+1] += digits[i]/MOD;
digits[i] %= MOD;
}
}
const bool operator<(Large& other)
{
if(sign != other.sign) return sign < other.sign;
if(sign == 0 && other.sign == 0) return false;
for(int i=LEN-1; i>=0; i--)
{
if(sign*digits[i] < sign*other.digits[i]) return true;
if(sign*digits[i] > sign*other.digits[i]) return false;
}
return false;
}
void setVeryLow()
{
sign = -1;
memset(digits, 31, sizeof digits);
}
void printOut()
{
if(sign == 0)
{
cout << 0 << endl;
return;
}
if(sign == -1)
cout << "-";
int GRV = -1;
for(int i=LEN-1; i>=0 && GRV == -1; i--)
if(digits[i] != 0)
GRV = i;
cout << digits[GRV];
for(int i=GRV-1; i>=0; i--)
cout << setfill('0') << setw(MS) << digits[i];
cout << endl;
}
};

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
while(cin >> arr[0])
{
int n = 1;
Large bestPreProd; bestPreProd.setVeryLow();
while(arr[n-1] != -999999) cin >> arr[n++];
n--;
for(int i=0; i<n; i++)
{
Large current; current = 1;
for(int j=i; j<n; j++)
{
current *= arr[j];
if(bestPreProd < current)
bestPreProd = current;
}
}
bestPreProd.printOut();
}
}

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