Project Euler Problem 36 Double-base palindromes

最新推荐文章于 2020-11-28 18:25:22 发布

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最新推荐文章于 2020-11-28 18:25:22 发布

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原文链接：http://www.cnblogs.com/tigerisland/p/7563991.html

本文介绍了一种算法，用于寻找并计算所有小于一百万且在十进制和二进制下均为回文数的整数之和。提供了两种C++实现方案，一种采用简单的进制转换和比较，另一种则利用字符数组进行中间处理。

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Double-base palindromes

Problem 36

The decimal number, 585 = 1001001001₂ (binary), is palindromic in both bases.

Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.

(Please note that the palindromic number, in either base, may not include leading zeros.)

C++(Simpler):

#include <iostream>

using namespace std;

const int N = 1000000;

bool ispalindrom(int n, int base)
{
    int palindrom = 0, temp;

    temp = n;
    while(temp) {
        palindrom *= base;
        palindrom += temp % base;
        temp /= base;
    }

    return n == palindrom;
}

int main()
{
    long total = 0;

    for(int i=1; i<N; i++) {
        if(!ispalindrom(i, 10))
            continue;

        if(!ispalindrom(i, 2))
            continue;

        total += i;
    }

    cout << total << endl;

    return 0;
}

C++:

#include <iostream>
#include <cstring>

using namespace std;

const int N = 1000000;

void myitoa(int n, char *p, int base)
{
    while(n) {
        *p++ = '0' + n % base;
        n /= base;
    }

    *p = '\0';
}

bool ispalindrom(char s[])
{
    int start = 0;
    int end = strlen(s) - 1;

    while(start < end) {
        if(s[end] != s[start])
            return false;
        start++;
        end--;

    }

    return true;
}

int main()
{
    char s[32];
    long total = 0;

    for(int i=1; i<N; i++) {
        myitoa(i, s, 10);
        if(!ispalindrom(s))
            continue;

        myitoa(i, s, 2);
        if(!ispalindrom(s))
            continue;

        total += i;
    }

    cout << total << endl;

    return 0;
}

转载于:https://www.cnblogs.com/tigerisland/p/7563991.html