练习题(1)-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_54669536/article/details/125817383

该程序主要计算不同金额范围内的累加税率，根据输入的整数金额判断适用的税率段并返回总税额。使用了switch-case结构和if-else语句实现，涵盖了5个税率段。

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#include<stdio.h>
#include<conio.h>
#include<assert.h>
#include<stdlib.h>
#include<math.h>


int main()
{
	int i;
	int sum;
	double sum1 = 100000 * 0.1;
	double sum2 = sum1+100000 * 0.075;
	double sum3 = sum2+200000 * 0.05;
	double sum4 = sum3+200000 * 0.03;
	double sum5 = sum4+400000 * 0.015;
	printf("please inter i:");
	scanf_s("%d", &i);
	int D = i / 100000;
	switch (D)
	{
	case 0:sum=100000 * 0.1;
		break;
	case 1:sum=sum1 + (i - 100000) * 0.075;
		break;
	case 2:sum = sum2 + (i - 200000) * 0.05;
		break;
	case 3:sum = sum3 + (i - 400000) * 0.03;
		break;
	case 4:sum = sum4 + (i - 600000) * 0.015;
		break;
	case 5:sum = sum5 + (i - 1000000) * 0.01;
		break;
	}
	printf("%d", sum);
	return 0;
}

int main()
{
	int sum = 0;
	int i;
	printf("please inter i:");
	scanf_s("%d", &i);
	double sum1 = 100000 * 0.1;
	double sum2 = sum1+100000 * 0.075;
	double sum3 = sum2+200000 * 0.05;
	double sum4 = sum3+200000 * 0.03;
	double sum5 = sum4+400000 * 0.015;
	if (i <= 100000)
	{
		sum = i * 0.1;
	}
	else if(i>100000&&i<=200000)
	{
		sum = sum1+(i - 100000) * 0.075;
	}
	else if(i>200000&&i<=400000)
	{
		sum = sum2 + (i - 200000) * 0.05;
	}
	else if (i > 400000 && i <= 600000)
	{
		sum =sum3+ (i - 400000) * 0.03;
	}
	else if (i > 600000 && i <= 1000000)
	{
		sum = sum4+ (i - 600000) * 0.015;
	}
	else if (i > 1000000)
	{
		sum =sum5 + (i - 1000000) * 0.01;
	}
	printf("%d", sum);
	return 0;
}
/*
int Printf1(int x)
{
	int num = 0;
	while (x != 0)
	{
		x = x / 10;
		num += 1;
	}
	return num;
}


void Printf2(int x)
{
	if (x != 0)
	{
		printf("%d", x % 10);
		printf("\n");
		Printf2(x / 10);
	}
}


void Printf3(int x)
{
	if(x != 0)
	{
		printf("%d", x % 10);
		Printf3(x / 10);
	}
}

int main()
{
	int a;
	printf("please inter a:");
	scanf_s("%d", &a);
	int sum=Printf1(a);
	printf("%d", sum);
	printf("\n");
	Printf2(a);
	printf("\n");
	Printf3(a);
	return 0;
}

int Printf(int x)
{
	if (x > 1000)
	{
		printf("请重新输入：");
	}
	else
	{
		return sqrt(x);
	}
}

int main()
{
	int a;
	printf("please inter a:");
	scanf_s("%d", &a);
	int tmp=Printf(a);
	printf("%d", tmp);
	return 0;
}