//copyright@ saturnman 
//updated@ 2011 July 
#include "stdafx.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 num1Ofn(int n)
{
int count=0;
while(n>0)
{
if(n%10 == 1)
count++;
n/=10;
}
return count;
}
int count1Num(int n)
{
int sum=0;
for (int i =1;i <= n;i ++)
{
sum+=num1Ofn(i);
}
return sum;
}
int main()
{
printf("%d\n", count1Num(21345));
printf("%d\n", NumberOf1BeforeBetween1AndN_Solution2(21345));
return 0;
}

本文转自博客园知识天地的博客，原文链接：十进制数1~n中1出现的次数，如需转载请自行联系原博主。