C语言程序设计教程（第三版）课后习题5.8 （C语言代码）

printf和scanf需要调用cstdio库吧

#include <stdio.h>
int main(void)
{
    int I;              //利润
    int ward;

    scanf("%d",&I);
    if(100000>=I)
        ward=I*0.1;
    if(100000<I&&I<=200000)
        ward=10000+(I-10000)*0.075;
        else if(200000<I&&I<=400000)
            ward=10000+100000*0.075+(I-200000)*0.05;
        else if(400000<I&&I<=600000)
            ward=10000+100000*0.075+200000*0.05+(I-400000)*0.03;
        else if(600000<I&&I<=1000000)
            ward=10000+100000*0.075+200000*0.05+200000*0.03+(I-600000)*0.015;
        else if(I>1000000)
            ward=10000+100000*0.075+200000*0.05+200000*0.03+400

#include <stdio.h>

int main()
{
	int I;
	int reward =0;
	scanf("%d",&I);
	int n =I;
	double a[6]={0.10,0.075,0.05,0.03,0.015,0.01};
	int b[6]={0,100000,200000,400000,600000,1000000};
	int i=0,j=0;

	while(n>b[j+1]-b[j]){
		reward += (b[j+1]-b[j])*a[j];
		n = I-b[j+1];
		j ++;
		if(j==5)
			break;
	}
	reward +=n*a[j];
	printf("%d\n",reward);
	
	return 0;
}

#include <stdio.h>

int main()
{
	int I;
	int reward =0;
	scanf("%d",&I);
	int n =I;
	double a[6]={0.10,0.075,0.05,0.03,0.015,0.01};
	int b[6]={0,100000,200000,400000,600000,1000000};
	int i=0,j=0;

	while(n>b[j+1]-b[j]){
		reward += (b[j+1]-b[j])*a[j];
		n = I-b[j+1];
		j ++;
		if(j==5)
			break;
	}
	reward +=n*a[j];
	printf("%d\n",reward);
	
	return 0;
}

#include<stdio.h>
int main()
{
	int a;
	int exact=0;
	int q=100000,w=200000,e=400000,r=600000,t=1000000;
	
	scanf("%d",&a);
	if(a>t)
    	{
			exact =(a-t)*0.01+(t-r)*0.015+(r-e)*0.03+(e-w)*0.05+(w-q)*0.075+q*0.1;
			printf("%d",exact);
 	    }
		else if(a>r&&a<=t)
			{
					exact =(a-r)*0.15+(r-e)*0.03+(e-w)*0.05+(w-q)*0.075+q*0.1;
					printf("%d",exact);
				}
			else if(a>e&&a<=r)
					{
						exact =(a-e)*0.03+(e-w)*0.05+(w-q)*0.075+q*0.1;
						printf("%d",exact);	
					}
				else if(a>w&&a<=e)
						{	 
							exact =(a-w)*0.05+(w-q)*0.075+q*0.1;
							printf("%d",exact);
						}

#include<stdio.h>
#define P 100000
int main()
{
     float i;
    scanf("%f",&i);
    if(i<0.0) printf("输入不合法");
    
    else if (i<P) printf("%f",i*0.1);
    else if(i>=P&&i<2*P) printf("%f",P*0.1+(i-P)*0.075);
    else if(i>=2*P&&i<4*P) printf("%f",P*0.1+P*0.075+(i-P)*0.05);
    else if(i>=P*4&&i<6*P) printf("%f",P*0.1+P*0.075+2*P*0.05+(i-P)*0.03);
    else if(i>=6*P&&i<10*P) printf("%f",P*0.1+P*0.075+2*P*0.05+2*P*0.03+(i-6*P)*0.015);
    else  printf("%f",P*0.1+P*0.075+2*P*0.05+2*P*0.03+4*P*0.015+(i-10*P)*0.01);
    
    return 0;
}

#include<stdio.h>
int main()
{
 int a,b,c,d,e,f,z;
scanf("%d",&z); 
a=z/10 ;
b=100000/10+(z-100000)*0.075;
c=(z-200000)*0.05+100000/10+100000*0.075;
d=100000/10+100000*0.075+200000*0.05+(z-400000)*0.03;
e=100000/10+100000*0.075+200000*0.05+200000*0.03+(z-600000)*0.015;
f=100000/10+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(z-1000000)*0.01;

if(z<=100000)
 {
	 printf("%d\n",a);
 
 }else if(z>100000&&z<=200000)
 {
	 printf("%d\n",b);
 }
 else if(z>200000&&z<=400000)
 {
	 printf("%d\n",c);
 }
 else if(z>400000&&z<=600000)
 {
	 printf("%d\n",d);
 }
 else if(z>600000&&z<=1000000)
 {
	 print

case 5: 和case 9: 少了break吧，虽然可以AC