Project Euler Problem 4: Largest palindrome product

最新推荐文章于 2021-06-26 21:21:40 发布

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最新推荐文章于 2021-06-26 21:21:40 发布

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原文链接：http://www.cnblogs.com/tigerisland/p/7564042.html

本文介绍了如何使用C++编程语言解决寻找两个三位数相乘得到的最大回文数的问题，并提供了两种不同的实现方法，最终得出906609为所求的最大回文数。

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Largest palindrome product

Problem 4

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.

C++:

#include <iostream>

using namespace std;

const int FROM = 100;
const int TO = 999;

bool ispalindrome(int product)
{
    int miror = 0, temp;

    temp = product;
    while(temp) {
        miror *= 10;
        miror += temp % 10;

        temp /= 10;
    }

    return miror == product;
}

int main()
{
    int maxpalindrome = 0, temp;

    for(int i=TO; i>=FROM; i--)
        for(int j=TO; j>=FROM; j--) {
            temp = i * j;
            if(temp < maxpalindrome)
                break;

            if(ispalindrome(temp))
                if(temp > maxpalindrome)
                    maxpalindrome = temp;
        }

    cout << maxpalindrome << endl;

    return 0;
}

C++:

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>

using namespace std;

const int MAXN = 1000000;
const int FROM = 100;
const int TO = 999;

struct node {
    int a, b, product;
    bool operator < (const node& n) const {
        return product < n.product;
    }
};

bool ispalindrome1(int product)
{
    char s[MAXN+1];
    int start, end;
    bool isp = true;

    sprintf(s, "%d", product);

    start = 0;
    end = strlen(s)-1;
    while(start <= end) {
        if(s[start] != s[end]) {
            isp = false;
            break;
        }
        start++;
        end--;
    }

    return isp;
}

bool ispalindrome2(int product)
{
    int miror = 0, temp;

    temp = product;
    while(temp) {
        miror *= 10;
        miror += temp % 10;

        temp /= 10;
    }

    return miror == product;
}

int main()
{
    priority_queue<node> q;
    node t;

    for(int i=FROM; i<=TO; i++)
        for(int j=FROM; j<=TO; j++) {
            t.a = i;
            t.b = j;
            t.product = i * j;

            q.push(t);
        }

    while(!q.empty()) {
        t = q.top();
        q.pop();

        if(ispalindrome2(t.product)) {
            cout << t.product << endl;
            break;
        }
    }

    return 0;
}

Run results:

906609

转载于:https://www.cnblogs.com/tigerisland/p/7564042.html