算法1 直接计数

由于a，b的大小都不大，所以可以直接枚举每个数的每一位，如果是2，就答案ans++

C++ 代码

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int ans,r;
int main()
{
    int a,b;
    cin>>a>>b;
    for(int i=a;i<=b;i++)
    {
        int t=i;
        while(t)
        {
            r=t%10;
            t/=10;
            if(r==2) ans++;
        }
    }
    cout<<ans;
    return 0;
}

算法2 数位Dp

Acwing338计数问题
数位Dp讲解

C++ 代码

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

const int N = 10;

int get(vector<int> num, int l, int r)
{
    int res = 0;
    for (int i = l; i >= r; i -- ) res = res * 10 + num[i];
    return res;
}

int power10(int x)
{
    int res = 1;
    while (x -- ) res *= 10;
    return res;
}

int count(int n, int x)
{
    if (!n) return 0;

    vector<int> num;
    while (n)
    {
        num.push_back(n % 10);
        n /= 10;
    }
    n = num.size();

    int res = 0;
    for (int i = n - 1 - !x; i >= 0; i -- )
    {
        if (i < n - 1)
        {
            res += get(num, n - 1, i + 1) * power10(i);
            if (!x) res -= power10(i);
        }

        if (num[i] == x) res += get(num, i - 1, 0) + 1;
        else if (num[i] > x) res += power10(i);
    }

    return res;
}

int main()
{
    int a,b;
    cin>>a>>b;
    cout << count(b, 2) - count(a - 1, 2) << ' ';
    return 0;
}