1010: [编程入门]利润计算

解题思路:利用case进行累加

注意事项:每次累加后，利润profit更新为被减数

参考代码:

#include <stdio.h>

int main()

{

int profit; //当月利润

int bonus=0; //奖金

scanf("%d",&profit);

if(profit>=0)

switch(profit/100000)

{

default:bonus+=(profit-1000000)*0.01;profit=1000000;

case 9:

case 8:

case 7:

case 6:bonus+=(profit-600000)*0.015; profit=600000;

case 5:

case 4:bonus+=(profit-400000)*0.03;  profit=400000;

case 3:

case 2:bonus+=(profit-200000)*0.05;  profit=200000;

case 1:bonus+=(profit-100000)*0.075; profit=100000;

case 0:bonus+=profit*0.1;

}

printf("%d",bonus);

}