解题思路:
先计算输入的数处于哪个挡位,然后根据挡位计算输出值,可以使代码更加简洁,并不需要写很多数放在上面,看着就会很头疼
注意事项:
后面乘以1000是为了保证算出来的数为整数,因为比例是小数,这里我用的为int型,所以范围支持没有很大,但是解决此题完全够用
参考代码:
#include <stdio.h>
int main(void){
int int_data = 0;
int profit_buff[6] = {0,100000,200000,400000,600000,1000000};
float per_buff[6] = {0.1f,0.075f,0.05f,0.03f,0.015f,0.01f};
short bit_count = 0;
int out_data = 0;
scanf("%d",&int_data);
for(;bit_count < 6;bit_count++){
if(int_data <= profit_buff[bit_count])
break;
}
out_data = (int_data-profit_buff[bit_count-1])*per_buff[bit_count-1]*1000;
for(;bit_count > 0;bit_count--){
if(bit_count-2 < 0)
break;
out_data += (profit_buff[bit_count-1]-profit_buff[bit_count-2])*per_buff[bit_count-2]*1000;
}
printf("%d",out_data/1000);
return 0;
}
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