C程序设计（第五版）谭浩强第三章课后练习题答案

#include <math.h>
#include <stdio.h>
#define PI 3.1415926
int lilv(){
    
    
	const int x=10;
	printf("%lf\n",pow(1+0.07,10));
	return 0;
} 


int cunkuan(){
    
    
	int x=1000;
	double s1,s2,s3,s4,s5;
	s1=x*(1+5*0.03);//五年存款 
	s2=(x*(1+0.021*2))*(1+0.0275*3);//两年三年存款 
	s3=(x*(1+0.0275*3))*(1+0.021*2);//三年两年存款 
	s4=x*pow((1+0.015),5);//一年年存款 
	s5=x*pow((1+0.0035),20);//季度存款 
	printf("s1:%lf元\n",s1);
	printf("s2:%lf元\n",s2);
	printf("s3:%lf元\n",s3);
	printf("s4:%lf元\n",s4);
	printf("s5:%lf元\n",s5);
	return 0;
}

int Daikuan(){
    
    
	int d=300000;
	int p=6000;
	double r=0.01; 
	double i=0;
	while(d>=0){
    
    
		i++;
		d=d*(1+r)-p;
	}
	printf("%lf\n",i);
	return 0;
}
void fenxi1(){
    
    
	char c1,c2;
	c1=97;
	c2=98;
	printf("c1=%c,c2=%c\n",c1,c2);
	printf("c1=%d,c2=%d\n",c1,c2);
} 
void Mima(){
    
    
	char c1,c2,c3,c4,c5;
	c1=getchar();
	c2=getchar();
	c3=getchar();
	c4=getchar();
	c5=getchar();
	c1+=4;c2+=4;c3+=4;c4+=4;c5+=4;
	putchar(c1);
	putchar(c2);
	putchar(c3);
	putchar(c4);
	putchar(c5);
	putchar('\n');
}
void yuan(){
    
    
	double zhouchang;
	double mianji;
	double qiubiaomianji;
	double qiutiji;
	double zhutiji;
	double r=1.5,h=3;
	zhouchang=PI*2*r;
	mianji=PI*r*r;
	qiubiaomianji=4*PI*r*r;
	qiutiji=3.0/4.0*PI*pow(r,3);
	zhutiji=PI*r*r*h; 
	printf("%lf \n%lf \n%lf \n%lf \n%lf \n",zhouchang,mianji,qiubiaomianji,qiutiji,zhutiji);
}
int main(){
    
    
	char ch;
	ch=getchar();
	switch(ch){
    
    
		case'1': lilv();
		break;
		case'2':cunkuan();
		break;
		case'3' : Daikuan();
		break;
		case'4':fenxi1();
		break;
		case'6':Mima();
		break;
		case'7':yuan();
		 break;
		 default:printf("ERROR!\n");
	}
	return 0;
} 

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