C :

#include<stdio.h>  
  
long long int Count(long long int n){  
    //1的个数  
    long long int count = 0;  
    //当前位  
    long long int Factor = 1;  
    //低位数字  
    long long int LowerNum = 0;  
    //当前位数字  
    long long int CurrNum = 0;  
    //高位数字  
    long long int HigherNum = 0;  
    if(n <= 0){  
        return 0;  
    }  
    while(n / Factor != 0){  
        //低位数字  
        LowerNum = n - (n / Factor) * Factor;  
        //当前位数字  
        CurrNum = (n / Factor) % 10;  
        //高位数字  
        HigherNum = n / (Factor * 10);  
        //如果为0,出现1的次数由高位决定  
        if(CurrNum == 0){  
            //等于高位数字 * 当前位数  
            count += HigherNum * Factor;  
        }  
        //如果为1,出现1的次数由高位和低位决定  
        else if(CurrNum == 1){  
            //高位数字 * 当前位数 + 低位数字 + 1  
            count += HigherNum * Factor + LowerNum + 1;  
        }  
        //如果大于1,出现1的次数由高位决定  
        else{  
            //（高位数字+1）* 当前位数  
            count += (HigherNum + 1) * Factor;  
        }  
        //前移一位  
        Factor *= 10;  
    }  
    return count;  
}  
  
int main(){  
    long long int a;  
    while(scanf("%lld",&a) != EOF){  
        printf("%lld\n",Count(a));  
    }  
    return 0;  
} 

C++ :

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int NumberOf1(const char* strN);
int PowerBase10(unsigned int n);

/////////////////////////////////////////////////////////////////////////////
// Find the number of 1 in an integer with radix 10
// Input: n - an integer
// Output: the number of 1 in n with radix
/////////////////////////////////////////////////////////////////////////////
int NumberOf1BeforeBetween1AndN_Solution2(int n)
{
      if(n <= 0)
            return 0;

      // convert the integer into a string
      char strN[50];
      sprintf(strN, "%d", n);

      return NumberOf1(strN);
}


/////////////////////////////////////////////////////////////////////////////
// Find the number of 1 in an integer with radix 10
// Input: strN - a string, which represents an integer
// Output: the number of 1 in n with radix
/////////////////////////////////////////////////////////////////////////////
int NumberOf1(const char* strN)
{
      if(!strN || *strN < '0' || *strN > '9' || *strN == '\0')
            return 0;

      int firstDigit = *strN - '0';
      unsigned int length = static_cast<unsigned int>(strlen(strN));

      // the integer contains only one digit
      if(length == 1 && firstDigit == 0)
            return 0;

      if(length == 1 && firstDigit > 0)
            return 1;

      // suppose the integer is 21345
      // numFirstDigit is the number of 1 of 10000-19999 due to the first digit
      int numFirstDigit = 0;
      // numOtherDigits is the number of 1 01346-21345 due to all digits
      // except the first one
      int numOtherDigits = firstDigit * (length - 1) * PowerBase10(length - 2);
      // numRecursive is the number of 1 of integer 1345
      int numRecursive = NumberOf1(strN + 1);

      // if the first digit is greater than 1, suppose in integer 21345
      // number of 1 due to the first digit is 10^4. It's 10000-19999
      if(firstDigit > 1)
            numFirstDigit = PowerBase10(length - 1);

      // if the first digit equals to 1, suppose in integer 12345
      // number of 1 due to the first digit is 2346. It's 10000-12345
      else if(firstDigit == 1)
            numFirstDigit = atoi(strN + 1) + 1;

      return numFirstDigit + numOtherDigits + numRecursive;
}

/////////////////////////////////////////////////////////////////////////////
// Calculate 10^n
/////////////////////////////////////////////////////////////////////////////
int PowerBase10(unsigned int n)
{
      int result = 1;
      for(unsigned int i = 0; i < n; ++ i)
            result *= 10;

      return result;
}


int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
		printf("%d\n",NumberOf1BeforeBetween1AndN_Solution2(n));
	return 0;
}