（算法）从0到n整数中数字2出现的次数

#include<iostream>
#include<sstream>
#include<math.h>
using namespace std;

template<class out_T,class in_T>
out_T convert(const in_T &t){
    stringstream ss;
    out_T result;
    ss<<t;
    ss>>result;
    return result;
}

int count2sInRangeAtDigit(int number,int d){
    int powerOf10=pow(10,d);
    int nextPowerOf10=powerOf10*10;
    int right=number%powerOf10;

    int roundDown=number-number%nextPowerOf10;
    int roundUp=roundDown+nextPowerOf10;

    int digit=(number/powerOf10)%10;
    if(digit<2)
        return roundDown/10;
    else if(digit==2)
        return roundDown/10+right+1;
    else
        return roundUp/10;
}

int count2sInRange(int number){
    int count=0;
    string str;
    str=convert<string>(number);
    int len=str.size();

    for(int digit=0;digit<len;digit++)
        count+=count2sInRangeAtDigit(number,digit);
    return count;
}

int main(){
    int n;
    while(cin>>n){
        cout<<count2sInRange(n)<<endl;
    }
    return 0;
}