lind5-8s.c

来自「谭浩强第二版的课后习题的答案的源程序」· C语言 代码 · 共 41 行

#include<stdio.h>
main()
	{
	double dividend,i,t;
	int grade=1;              //dividend means "hong li, ti cheng",i represent benifet
	printf("\nPlease input the benefit.\n");
	scanf("%lf",&i);
	t=i;
	i=i-((long)i%100000);
	grade+=(long)i/1e5;
	if(grade>0)
	{
		switch(grade)
		{
			case 1:
			dividend=t*0.1;break;
			case 2:
			dividend=1e4+(t-1e5)*.075;break;
			case 3:
			case 4:
			dividend=1e4+1e5*.075+(t-2e5)*.05;break;
			case 5:
			case 6:
			dividend=1e4+1e5*.075+2e5*.05+(t-4e5)*.03;break;
			case 7:
			case 8:
			case 9:
			case 10:
			dividend=1e4+1e5*.075+2e5*.05+2e5*.03+(t-6e5)*.015;break;
			default:
			dividend=1e4+1e5*.075+2e5*.05+2e5*.03+4e5*.015+(t-1e6)*.01;break;
			printf("The dididend is %.2fyuan.\n",dividend);
		}
	printf("The dididend is %.2fyuan.\n",dividend);
	}
	else
	printf("It is a false date.\n");
	}