找1到n所有整数出现1的个数

编程之美2.4

n=12时，1，11，12这3个数包含1，所以1的个数是5.

Line 9是为了防止factor溢出。

 1 #include <iostream>

 2 #include <stdlib.h>

 3 #include <time.h>

 4 using namespace std;

 5 

 6 int countOne(int n) {

 7     int ans = 0;

 8     int factor = 1;

 9     while (n / factor >= 10) { // avoid overflow

10         int high = n / factor / 10;

11         int low = n % factor;

12         int current = (n / factor) % 10;

13         if (current == 0) {

14             ans += high * factor;

15         } else if (current == 1) {

16             ans += high * factor + low + 1;

17         } else {

18             ans += (high + 1) * factor;

19         }

20         factor *= 10;

21     }

22     if (n / factor == 1) ans += (n % factor) + 1;

23     else if (n / factor != 0) ans += factor;

24     return ans;

25 }

26 

27 int bruteForce(int n) {

28     int ans = 0;

29     for (int i = 1; i <= n; ++i) {

30         int tmp = i;

31         while (tmp) {

32             if (tmp % 10 == 1) ans++;

33             tmp /= 10;

34         }

35     }

36     return ans;

37 }

38 

39 int main() {

40     srand(time(NULL));

41 

42     for (int i = 0; i < 100; ++i) {

43         int n = rand() % 9999999;

44         int c1 = countOne(n);

45         int c2 = bruteForce(n);

46         if (c1 != c2) {

47             cout << c1 << " " << c2 << " " << n << endl;

48         }

49     }

50     return 0;

51 }