Algorithm --> 小于N的正整数含有1的个数

//https://leetcode.com/problems/number-of-digit-one/

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

class Solution {
public:
    int countDigitOne(int n) {
        int ones = 0;
        for (long long m = 1; m <= n; m *= 10)
            ones += (n/m + 8) / 10 * m + (n/m % 10 == 1) * (n%m + 1);
        return ones;       
    }
};

 

转载于:https://www.cnblogs.com/jeakeven/p/4737797.html