//程序格式(输入：long long范围内的正整数；输出：0~9每位数字出现的次数，以空格分隔）
//数字0~9出现次数,思路来自《编程之美》计算数字1的个数，拓展之#include <iostream>#include <vector>//#include <bits/stdc++.h>using namespace std;long long counteach(long long n,int i)  //统计数字i出现个数{    long long iCount=0;    long long iFactor   =1;    long long iLowerNum =0;    long long iCurrNum  =0;    long long iHigherNum=0;    while(n/iFactor !=0)    {        iLowerNum = n-(n/iFactor)*iFactor;        iCurrNum  = (n/iFactor)%10;        iHigherNum= n/(iFactor*10);        if(iCurrNum<i)            iCount += iHigherNum*iFactor;        else if(i==iCurrNum)            iCount += iHigherNum*iFactor+iLowerNum+1;        else if(iCurrNum>i)            iCount += (iHigherNum+1)*iFactor;        //处理0的个数        //若n为1位数，比如本来是1 2 3 4 5 6 ，之前处理成 0 1 2 3 4 5 6,多加了1个0        //若n为2位数，比如本来是1 2 3 4 5 6 7 8 9 10 11 12，之前处理成 00 01 02 ...09 10 11 12,多加了1+10个0        //若n为3位数，比如本来是1 2 3 4 ... 115，之前处理成000 001 002 ...009 010 011...099 100...115，多加了1+10+100个0        //因此需要在每层循环中减去多计算的0的个数        if(0==i)            iCount-=iFactor;        iFactor*=10;    }    return iCount;}vector<long long> count(long long n)    //依次统计0~9{    vector<long long>res(10,0);    for(int i=0;i<10;i++)    {        res[i]+=counteach(n,i);    }   return res;}int main(){    long long n;    cin>>n;    vector<long long>res=count(n);    for(int i = 0; i < res.size(); i++)    {        i == 0 ? cout << res[i] : cout << " " << res[i];    }    return 0;}