（Problem 36）Double-base palindromes

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.)

题目大意：

十进制数字585 = 1001001001₂ (二进制),可以看出在十进制和二进制下都是回文（从左向右读和从右向左读都一样）。

求100万以下所有在十进制和二进制下都是回文的数字之和。

（注意在两种进制下的数字都不包括最前面的0）

//（Problem 36）Double-base palindromes
// Completed on Thu, 31 Oct 2013, 13:12
// Language: C
//
// 版权所有（C）acutus   (mail: acutus@126.com) 
// 博客地址：http://www.cnblogs.com/acutus/
#include<stdio.h>
#include<stdbool.h>

bool test(int *a, int n)
{
    bool flag = true;
    for(int i = 0; i < n/2; i++) {
        if(a[i] != a[n-i-1]) {
            flag = false;
            break;
        }
    }
    return flag;
}

bool palindromes(int n, int base)  //判断整数n在基为base时是否为回文数
{
    int a[100];
    int i = 0;
    while(n) {
        a[i++] = n % base;
        n /= base;
    }
    return test(a,i);
}

int main(void)
{
    int sum = 0;
    for(int i = 1; i <= 1000000; i += 2)
    {
        if(palindromes(i, 10) && palindromes(i, 2))
            sum += i;
    }
    printf("%d
", sum);
    return 0;
}

Answer:

872187