The task is simple: given any positive integer N, you are supposed to count the total number of 1’s in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1’s in 1, 10, 11, and 12.

Input Specification:

Each input file contains one test case which gives the positive N (≤230).

Output Specification:

For each test case, print the number of 1’s in one line.

Sample Input:

12

Sample Output:

5

思路

从低位到高位考虑每一位为 1 的个数，设当前位的数为 i，该位左边的数为 leftNum，右边的数为 rightNum，e 从 1 开始每次遍历乘 10。通过数字举例，可以总结出这样的公式：

1. i > 1 时，个数为 (leftNum + 1) * e
2. i == 1 时，个数为 leftNum * e + rightNum + 1
3. i == 0 时，个数为 leftNum * e

#include <cstdio>

int main() {
    int n;
    scanf("%d", &n);
    int i, leftNum = n, rightNum, e = 1, ans = 0;
    while (leftNum != 0) {
        i = n / e % 10;
        leftNum /= 10;
        rightNum = n % e;
        if (i > 1) {
            ans += (leftNum + 1) * e;
        } else if (i == 1) {
            ans += leftNum * e + rightNum + 1;
        } else {
            ans += leftNum * e;
        }
        e *= 10;
    }
    printf("%d", ans);
    return 0;
}