100以内含九的个数计算

int main()
{
int a = 0,count = 0;
for (a = 1; a <= 100; a++)
{
if (a % 10 == 9)//余9的数
{
count++;
}
if (a / 10 == 9)//商9的数
{
count++;
}
}
printf("\ncount = %d\n", count);
return 0;
}