#include <stdio.h>

//整数1到n，字符ch出现的次数；如1到12，1出现5次

int count1s (int n, char ch){

	int i = ch - '0';

	int count = 0;

	int j = 1;

	int m = n;

	int t = 0;

	int k = 1;

	if (i == 0) i = 10;

	
　　　　　//循环比较最后一位，/10 移位

	while (m > 0){

		int t = i;

		int count1 = 0;

		while (t <= m){

			count1++;

			t += 10;

		}

		count += count1 * j;

		j *= 10;

		m /= 10;

	}

	//减去多加的部分; 如110中计算1出现个数，在计算十位时多加9，需要减去。在计算百位是也多加了89，需要减去

	m = n;

	j = 0;

	while (m){

		if (m % 10 == ch - '0'){

			j = t;

			if (j != 0){

				int a = 0;

				k = 1;

				for (a = 0; a < j; a++) k *= 10;

					count = count - (k - (n % k) - 1);

			}

		}

		m /= 10;

		t++;

	}

	return count;

}

#if 1 //普通的逐个整数比较的测试程序

//某个整数中1出现的次数

int NumberOfCh(unsigned int n, char ch)

{

    int number = 0;

    while (n)

    {

        if(n % 10 == ch - '0')

            number ++;

        n = n / 10;

    }  

    return number;

}  

//求1到n中所有整数中1出现的总次数

int ceshi(unsigned int n, char ch)

{

    int number = 0;

	  int i = 1;

    for(i = 1; i <= n; ++ i)

        number += NumberOfCh(i, ch);   

    return number;

}

#endif

int main (void){

#if 0

	int i = 0;

	char ch = '0';

	for (i = 0; i < 65535; i++){

		for (ch = '0'; ch <= '9'; ch++){

			if (count1s (i, ch) != ceshi (i, ch)){

				printf ("%d %c\n", i, ch);

				return -1;

			}

		}

		printf ("%d\n", i);

	}

#else

	printf ("%d\n", count1s(1365535, '5'));

	printf ("%d\n", ceshi (1365535, '5'));

#endif

}