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Exponentiation by Squaring (Project Euler 99) Tips on my solution

开发者 https://www.devze.com 2023-03-28 23:17 出处:网络
Here is the problem I am talking about http://projecteuler.net/index.php?section=problems&id=99 My code will compile and run correctly. I am guessing the computation is where it is messing up. It

Here is the problem I am talking about http://projecteuler.net/index.php?section=problems&id=99

My code will compile and run correctly. I am guessing the computation is where it is messing up. It is telling me that line number 633 is the largest (which project euler says is incorrect).

#include <iostream>
#include <string>
#include <fstream>

using namespace std;

int poww(int base, int exp);
int main()
{
    //ignore messy/unused variables. I am desperate 
    int lineNumber = 0;
    string line; 
    int answerLine = 0;
    int max =0;
    int lineNum = 0;
    int answer =0;
    ifstream inFile;
    size_t location;
    string temp1,temp2;
    int tempMax = 0;
    int base,exp = 0;
    inFile.open("C:\\Users\\myYser\\Desktop\\base_exp.txt");
    while(getline(inFile,line))
    {
        lineNumber++;
        location = line.find(",");
        temp1 = line.substr(0,(int(location)));
        temp2 = line.substr((int(location)+1),line.length());
        //cout << temp1 << " " << temp2 << endl;
        base = atoi(temp1.c_str());
        exp =  atoi(temp2.c_str());
        tempMax= poww(base,exp);

        if (tempMax > max){
            max = tempMax;
            answer = base;
            answerLine = lineNumber;
        }

    }


    cout << answer << " " << answerLine;

    cin.get();
    return 0;
}
int poww(int base, int exp)
{
    int result = 1;
    while (exp)
    {
        if (exp & 1)
      开发者_开发技巧      result *= base;
        exp >>= 1;
        base *= base;
    }

    return result;
}


You're under-thinking this problem.

You need to come up with a way to scale down these numbers drastically so you can still compare them. In other words, you may want to look into a way of comparing how many digits the result will be.

A hint would be log(a^b) = b * log(a)


A 32-bit int can only hold 2^32 values, and some of those magically turn negative at some point...

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