Question

c语言谭浩强第四版 第二章课后题3.5.7

Answer 1

#include <stdio.h>
#include <math.h>
int main()
{float d=300000,p=6000,r=0.01,m;
m=log10(p/(p-d*r))/log10(1+r);
printf("m=%6.2f\n",m);
return 0;
}

#include <stdio.h>
int main()
{int a,b;
float x,y;
char c1,c2;
scanf("a=%d b=%d",&a,&b);
scanf("%f %e",&x,&y);
scanf("%c%c",&c1,&c2);
printf("a=%d,b=%d,x=%f,y=%f,c1=%c,c2=%c\n",a,b,x,y,c1,c2);
return 0;
}

#include <stdio.h>
int main ()
{float h,r,l,s,sq,vq,vz;
float pi=3.141526;
printf("请输入圆半径r，圆柱高h∶");
scanf("%f,%f",&r,&h); //要求输入圆半径r和圆柱高h
l=2*pi*r; //计算圆周长l
s=r*r*pi; //计算圆面积s
sq=4*pi*r*r; //计算圆球表面积sq
vq=3.0/4.0*pi*r*r*r; //计算圆球体积vq
vz=pi*r*r*h; //计算圆柱体积vz
printf("圆周长为: l=%6.2f\n",l);
printf("圆面积为: s=%6.2f\n",s);
printf("圆球表面积为: sq=%6.2f\n",sq);
printf("圆球体积为: v=%6.2f\n",vq);
printf("圆柱体积为: vz=%6.2f\n",vz);
return 0;
}