（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.)

 1 #include<stdio.h>

 2 #include<math.h>

 3 #include<stdbool.h>

 4 

 5 bool test(int *a, int n)

 6 {

 7     bool flag = true;

 8     for(int i = 0; i < n/2; i++) {

 9         if(a[i] != a[n-i-1]) {

10             flag = false;

11             break;

12         }

13     }

14     return flag;

15 }

16 

17 bool palindromes(int n, int base)  //判断整数n在基为base时是否为回文数

18 {

19     int a[100];

20     int i = 0;

21     while(n) {

22         a[i++] = n % base;

23         n /= base;

24     }

25     return test(a,i);

26 }

27 

28 int main(void)

29 {

30     int sum = 0;

31     for(int i = 1; i <= 1000000; i++)

32     {

33         if(palindromes(i, 10) && palindromes(i, 2))

34             sum += i;

35     }

36     printf("%d\n", sum);

37     return 0;

38 }

C代码

 下面是根据garbageMan兄所提出的优化方法（具体解释详见本文留言）修改后的代码：

 1 #include<stdio.h>

 2 #include<stdbool.h>

 3 

 4 bool test(int *a, int n)

 5 {

 6     bool flag = true;

 7     for(int i = 0; i < n/2; i++) {

 8         if(a[i] != a[n-i-1]) {

 9             flag = false;

10             break;

11         }

12     }

13     return flag;

14 }

15 

16 bool palindromes(int n, int base)  //判断整数n在基为base时是否为回文数

17 {

18     int a[100];

19     int i = 0;

20     while(n) {

21         a[i++] = n % base;

22         n /= base;

23     }

24     return test(a,i);

25 }

26 

27 int main(void)

28 {

29     int sum = 0;

30     for(int i = 1; i <= 1000000; i += 2)

31     {

32         if(palindromes(i, 10) && palindromes(i, 2))

33             sum += i;

34     }

35     printf("%d\n", sum);

36     return 0;

37 }

C代码

谢谢garbageMan兄的指正，大家还有什么优化方法可以留言，一起交流

 

Answer:

872187