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 }