View Code of Problem 70

#include<stdio.h>
int main(void)
{
	long m;
	int c,p;
	while(scanf("%ld",&m)!=EOF)
	{
		if(m>=1000000)
		{
			c=10;
		}
		else 
		   c = m/100000;
	    switch(c)
	   {
		 case 0 : p=m*0.1; break;
		 case 1 : p=10000+(m-100000)*0.075; break;
		 case 2 :  
		 case 3 : p=17500+(m-200000)*0.05; break;   //case 2，3是一种计算方法
		 case 4 : 
		 case 5 : p=27500+(m-400000)*0.03; break; 
		 case 6 :
		 case 7 :
		 case 8 :
		 case 9 : p=33500+(m-600000)*0.015; break; 
		 case 10: p=39500+(m-1000000)*0.01; break;     //注意case后的形式
	   }
	   printf("%d\n",p);
	}	
}