Algorithm questions daily practice---Day 72: The number of number 1

Get into the habit of writing together! This is the 7th day of my participation in the "Nuggets Daily New Plan · April Update Challenge", click to view the details of the event .

1. Problem description

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

Topic link: The number of number 1

Second, the subject requirements

Example 1

输入： n = 13
输出： 6
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Example 2

输入： n = 0
输出： 0
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visit

1.数学思想、超时优化
2.建议用时15~35min
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3. Problem Analysis

As soon as I got my hands on this question, I felt that something was wrong. It was a very simple question that only needed to be counted in a loop. Likou actually added a difficult label to it.

class Solution {
public:
    int countDigitOne(int n) {
        int i,k,ans=0;//初始化数据
        for(i=1;i<=n;i++)//循环1~n
        {
            k=i;
            while(k)//数位中是否包含1
            {
                if(k%10==1)
                    ans++;//计数
                k=k/10;
            }
        }
        return ans;//输出结果
    }
};
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So, my code definitely times out.

After reading the solution >_<, I found that this is a regular question:

The rules are as follows, drawing on the ideas of the big guys:

/**
从低往高位，计算每一位的数量：
第1位上的1的个数：1+(n-1)/10
第2位上的1的个数：(1+(n/10-1)/10)*10
第3位上的1的个数：(1+(n/100-1)/10)*100
……
第k+1位上的1的个数：(1+(n/(10^k)-1)/10)*(10^k)
如果n的第k+1位上是1，则说明可能没有填满该位，计算第k+1位的数量时还要 -10^k+1+n%(10^k)，相当于独立计算
 */
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Fourth, the encoding implementation

class Solution {
public:
    int countDigitOne(int n) {
        long long k = 1,ans = 0;//初始化数据
        while (n >= k){
            ans += (1+(n/k-1)/10)*k;//计数
            if (n/k%10 == 1) ans = ans-k+1+n%k;//当前位为1，减去数据
            k *= 10;
        }
        return ans;//输出结果
    }
};
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5. Test results