//计算出1-n(n通过传参实现)中有多少个9。
#include<stdio.h>
int main()
{
int n;
printf("Please input a number:\n");
scanf("%d",&n);
fun(n);
return 0;
}
void fun(int n)
int i,a,b;
int count = 0;
for(i = 1;i <= n;i++)
{
a = i;
while(a != 0)
{
b = a % 10;
a = a / 10;
if(b == 9)
{
count++;
}
}
}
printf("9的个数是:%d\n",count);
}原文:http://www.cnblogs.com/Janskid/p/7147602.html