要求是在 I<= 100000时，i=I*0.1，100000< I <= 200000时，i = 100000 * 0.1 + (I - 100000) * 0.075，200000< I <= 400000时，i = 100000 * 0.1 + 100000 * 0.075 + (I - 200000) * 0.05，400000< I <= 600000时，i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + (I - 400000) * 0.03，600000<I <= 1000000时，i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + (I - 600000)*0.015，用switch语句来做。
我做这个有问题，如果大家有更好的办法，或者能改正这个都行，谢谢了

#include<stdio.h>
void main()
{
int I, x;
float i;
printf("Please enter the profit: \n");
scanf("%d", &I);
x = I/100000;
switch(x)
{
case'0': printf("The praise of the year is %f\n", i = I * 0.1);
break;
case'1': printf("The praise of the year is %f\n", i = 100000 * 0.1 + (I - 100000) * 0.075);
break;
case'2': printf("The praise of the year is %f\n", i = 100000 * 0.1 + 100000 * 0.075 + (I - 200000) * 0.05);
break;
case'3': printf("The praise of the year is %f\n", i = 100000 * 0.1 + 100000 * 0.075 + (I - 200000) * 0.05);
break;
case'4': printf("The praise of the year is %f\n", i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + (I - 400000) * 0.03);
break;
case'5': printf("The praise of the year is %f\n", i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + (I - 400000) * 0.03);
break;
case'6': printf("The praise of the year is %f\n", i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + (I - 600000)*0.015);
break;
case'7': printf("The praise of the year is %f\n", i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + (I - 600000)*0.015);
break;
case'8': printf("The praise of the year is %f\n", i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + (I - 600000)*0.015);
break;
case'9': printf("The praise of the year is %f\n", i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + (I - 600000)*0.015);
break;
case'10': printf("The praise of the year is %f\n", i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + (I - 600000)*0.015);
break;
}
}

case 0 不是case '0'
后者表示的是0的ASCII码,好象是48

恩，我弄错了，现在改成这样的，应该要简洁得多。
#include<stdio.h>
void main()
{
int I, x;
float i;
printf("Please enter the profit: \n");
scanf("%d", &I);
x = (I - 1)/100000;
switch(x)
{
case 0: i = I * 0.1;
break;
case 1: i = 100000 * 0.1 + (I - 100000) * 0.075;
break;
case 2:
case 3: i = 100000 * 0.1 + 100000 * 0.075 + (I - 200000) * 0.05;
break;
case 4:
case 5: i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + (I - 400000) * 0.03;
break;
case 6:
case 7:
case 8:
case 9: i = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + (I - 600000)*0.015;
break;
}
printf("The praise of the year is %f\n", i);
}

[此贴子已经被作者于2007-2-4 12:29:57编辑过]