Project Euler 99:Largest exponential 最大的幂

Largest exponential

Comparing two numbers written in index form like 211 and 37 is not difficult, as any calculator would confirm that 211 = 2048 < 37 = 2187.

However, confirming that 632382518061 > 519432525806 would be much more difficult, as both numbers contain over three million digits.

Using base_exp.txt(right click and ‘Save Link/Target As…’), a 22K text file containing one thousand lines with a base/exponent pair on each line, determine which line number has the greatest numerical value.

NOTE: The first two lines in the file represent the numbers in the example given above.

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最大的幂

比较两个如211和37这样写成幂的形式的数并不困难，任何计算器都能验证211 = 2048 < 37 = 2187。

然而，想要验证632382518061 > 519432525806就会变得非常困难，因为这两个数都包含有超过三百万位数字。

22K的文本文件base_exp.txt（右击并选择“目标另存为……”）有一千行，每一行有一对底数和指数，找出哪一行给出的幂的值最大。

注意：文件的前两行就是上述两个例子。

解题

指数运算太大了，取对数不就可以了

百度百科找的图片。

原函数和其反函数关于y=x对称，并且单调性一样。

JAVA

package Level3;

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Set;
import java.util.TreeSet;

public class PE099{
    public static void run(){
        ArrayList<ArrayList<Integer>> base_exp = base_exp();
        int size = base_exp.size();
        double result[] = new double[size];
        double MAX = 1.0*Integer.MIN_VALUE;
        int index = 0;
        for(int i =0;i<size;i++){
            ArrayList<Integer> bp = base_exp.get(i);
            int base = bp.get(0);
            int exp = bp.get(1);
            result[i] = log10(base,exp);
//            System.out.println(result[i]);
            if(MAX < result[i]){
                MAX = result[i];
                index = i;
            }
        }
        // 要加一，你懂的
        index+=1;
        System.out.println(index);
    }
//    709
//    running time=0s22ms
    public static double log10(int base,int exp){
        double res = 0.0;
        res = exp*Math.log10(base);
        return res;
    }
    public static ArrayList<ArrayList<Integer>> base_exp(){
        String filename = "src/Level3/p099_base_exp.txt";
        ArrayList<ArrayList<Integer>> base_exp = new ArrayList<ArrayList<Integer>>();

        try {
            BufferedReader input = new BufferedReader(new FileReader(filename));
            String str="";
            try {
                while((str=input.readLine())!=null){
                    String[] strArr = str.split(",");
                    ArrayList<Integer> num = new ArrayList<Integer>();
                    num.add(Integer.parseInt(strArr[0]));
                    num.add(Integer.parseInt(strArr[1]));
                    base_exp.add(num);
                }
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        } catch (FileNotFoundException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return base_exp;

    }
    public static void main(String[] args) throws IOException{
        long t0 = System.currentTimeMillis();
        run();
        long t1 = System.currentTimeMillis();
        long t = t1 - t0;
        System.out.println("running time="+t/1000+"s"+t%1000+"ms");

    }
}

Python

# coding=gbk

import time as time
import re
import math
def run():
    filename = 'E:/java/projecteuler/src/Level3/p099_base_exp.txt'
    file = open(filename)
    MAX = 0.0
    index = 0
    i = 0
    for row in file.readlines():
        row = row.strip('\n').split(",")
        res = int(row[1])*math.log(int(row[0]))
        i+=1
        if res>MAX:
            MAX = res
            index = i
    print index
#
# running time= 0.00400018692017 s
t0 = time.time()
run()
t1 = time.time()
print "running time=",(t1-t0),"s"