projecteuler.net 4--Largest palindrome product

fs1237

于 2017-10-21 09:00:12 发布

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分类专栏：算法题文章标签：基础算法题

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本文介绍了一种通过编程方法寻找由两个三位数相乘得到的最大回文数的技术。使用C++实现了一个检查数字是否为回文数的函数，并通过遍历所有可能的三位数组合来找到符合条件的最大回文数及其因数。

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.

Find the largest palindrome made from the product of two 3-digit numbers.

#include <iostream>
#include <cmath>
bool check(int target)
{

    //std::cout<<target<<std::endl;
    int i=6;
    int j=0;
    int temp=0;
    int out=0;
    int hehe;
    hehe=target;
    if ((int)pow(10, i)>target)
    {

        //std::cout<<pow(10,i)<<std::endl;
        i--;
    }
    while(target!=0)
    {


            temp = target /(int)(pow(10, i));
            out=out+temp*pow(10,j);
            //std::cout<<out<<std::endl;
            target=target-temp*pow(10,i);
            i--;
            j++;

    }
    //std::cout<<out<<std::endl;
    if(out==hehe)
    {
        std::cout<<out<<" YES"<<std::endl;
        return (true);
    }
    else
        return(false);
}

int main2()
{
    //check(12345)==true;
    long sum,temp;
    int i,j;
    i=j=999;
    temp=0;
    int tempx,tempy;
    for(i=100;i<=999;i++)
    {
        for(j=100;j<=999;j++)
        {
            //std::cout<<i<<" "<<j<<std::endl;
            sum=i*j;
            if(check(sum)==true)
            {
                if(sum>=temp)
                {
                    temp = sum;
                    tempx=i;
                    tempy=j;
                }
            }
        }
    }
    std::cout<<temp<<" "<<tempx<<" "<<tempy<<" "<<std::endl;
}
int main()
{
    //check(906609);
    main2();
}