C语言程序设计教程（第三版）课后习题5.8 （C语言代码）

#include<stdio.h>
int main()
{
    int a,t;
    scanf("%d",&a);
    switch(a/100000)
    {
        case 0:t=a*0.1;break;
        case 1:t=100000*0.1+(a-100000)*0.075;break;
        case 2:
        case 3:t=100000*0.1+100000*0.075+(a-200000)*0.05;break;
        case 4：
        case 5:t=100000*0.1+100000*0.075+200000*0.05+(a-400000)*0.03;break;
        case 6:
        case 7:
        case 8：
        case 9:t=100000*0.01+100000*0.075+200000*0.05+200000*0.03+(a-600000)*0.015;break;
        default:t=100000*0.01+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(a-1000000)*0.01;break;
    }
    pri

int x,y,z,n;
    while(scanf("%d",&x)==1)
    {
        if(x<0)
        {
            printf("try again\n");
            continue;
        }
        y=0;
        n=0;
        z=100000;
        switch(x/z)                            
        {                                      
            default:y+=(n=(x-10*z-n))*0.01;    
            case 10:y+=(n=(x-10*z-n))*0.01;    
            case 6 :y+=(n=(x-6*z-n))*0.15;     
            case 4 :y+=(n=(x-4*z-n))*0.03;     
            case 2 :y+=(n=(x-2*z-n))*0.05;     
            case 1 :y+=(n=(x-z-n))*0.075;      
            case 0 :y+=(x-n

#include<stdio.h>
int main()
{
    int x,y,z,n;
    while(scanf("%d",&x)==1)
    {
        if(x<0)
        {
            printf("try again\n");
            continue;
        }
        y=0;
        n=0;
        z=100000;
        switch(x/z)                            //利用标签内没有break语句的话便会依次执行
        {                                      //每次计算出相应部分的奖金然后相加起来
            default:y+=(n=(x-10*z-n))*0.01;    //超出110w以上
            case 10:y+=(n=(x-10*z-n))*0.01;    //超出100w
            case 6 :y+=(n=(x-6*z-n))*0.15;     //超出60w
            case 4 :y+=(n=(x-4*z-n))*0.03;     //超出40w
            case

#include<stdio.h>
int main()
{
	int l,l1,y;
	scanf("%d",&l);
	l1=l/100000;
	if(l1<=1)
	  printf("%.0f",0.1*l); 
	else   
	switch(l1)
	{
	case 1:printf("%.0f",10000+0.075*(l-100000)); break;
	case 3:
	case 2:printf("%.0f",10000+7500+0.05*(l-200000)); break;
	case 5:
	case 4:printf("%.0f",10000+7500+10000+0.03*(l-400000)); break;
	case 7:
	case 6:
	case 8:
	case 9:printf("%.0f",10000+7500+10000+6000+0.015*(l-600000)); break;
	default :printf("%.0f",39500+0.01*(l-1000000)); 	
    }
	return 0;
}

#include<stdio.h>
int main()
{
    //获取输入奖金
    int a,s;
    scanf("%d",&a);
    if (a<=100000)
        s=a*0.1;
    if (a>100000 && a<=200000)
    {
        s=(a-1000000)*0.075+100000*0.1;
    }
    if (a>200000 && a<=400000)
    {
        s=(a-200000)*0.05+100000*(0.1+0.075);
    }
    if (a>400000 && a<=600000)
    {
        s=(a-400000)*0.03+100000*(0.1+0.075)+200000*0.05;
    }
    if (a>600000 && a<=1000000)
    {
        s=(a-600000)*0.015+100000*(0.1+0.075)+200000*(0.05+0.3);
    }
    if (a>1000000)
    {
        s=(a-1000000)*0.01+100000*(0.1+0.075)+200000*(0.05+0.03)+400000*0.015;

不明白为什么case 5的语句后不用加break，不加的话不是case 9的t覆盖了case 5的t了吗

#include<stdio.h>
int main()
{   int a,c,b;
    scanf("%d",&a);
    if(a>0)
    {
    	c=a*0.1;
    	b=100000*0.1;
    	if(a>100000)
    	{   
		    c=b+(a-100000)*0.075;
		    b=b+100000*0.075;  		
    		if(a>200000)
			{   
			    c=b+(a-200000)*0.05;
				b=b+200000*0.05;
				if(a>400000)
				{
				    c=b+(a-400000)*0.03;
					b=b+200000*0.03;
					if(a>600000)
					{
						c=b+(a-600000)*0.015;
						b=b+400000*0.015;
						if(a>1000000)
	                     c=b+(a-1000000)*0.01; 
					}					
				 } 
			 } 
		}
	}
	printf("%d",c);
	return 0;
}

#include<stdio.h>
int main()
{
	int a,b;
	printf("请输入当月利润:\n");
	scanf("%d",&a);
	if(a<=100000)
	{
		b=a*0.1;
	}
	else if(100000<a&&a<=200000)
	{
		b=10000+(a-100000)*0.075;
	}
	else if(200000<a&&a<=400000)
	{
		b=10000+7500+(a-200000)*0.05;
	}
	else if(400000<a&&a<=600000)
	{
		b=10000+7500+10000+(a-400000)*0.03;
	}
	else if(600000<a&&a<=1000000)
	{
		b=10000+7500+10000+6000+(a-600000)*0.015;
	}
	else if(a>1000000)
	{
		b=10000+7500+10000+6000+6000+(a-1000000)*0.01;
	}
	printf("应发的奖金数为：");
	printf("%d",b);
	return 0;
}