1 条题解

  • 0
    @ 2025-4-7 21:29:26

    C :

    #include<stdio.h>
    int main()
    {
    	float r,h;
    	float C1,Sa,Sb,Va,Vb;
    	while(scanf("%f%f",&r,&h)!=EOF)
    	{
    		C1=0;
    		Sa=0;
    		Sb=0;
    		Va=0;
    		Vb=0;
    		C1=2*3.14*r;
    		Sa=3.14*r*r;
    		Sb=4*3.14*r*r;
    		Va=(4.0/3)*3.14*r*r*r;
    		Vb=Sa*h;
    		printf("C1=%.2f\n",C1);
    		printf("Sa=%.2f\n",Sa);
    		printf("Sb=%.2f\n",Sb);
    		printf("Va=%.2f\n",Va);
    		printf("Vb=%.2f\n",Vb);
    	}
    	return 0;
    }
    
    

    C++ :

    #include<iostream>
    #include<cstdio>
    using namespace std;
    const double PI = 3.14;
    int main()
    {
    	float r,h;
    	cin>>r>>h;
    	float C1=2*PI*r;
    	float Sa=PI*r*r;
    	float Sb=4*PI*r*r;
    	float Va=4.0/3*PI*r*r*r;
    	float Vb=Sa*h;
    	printf("C1=%.2f\nSa=%.2f\nSb=%.2f\n",C1,Sa,Sb);
    	printf("Va=%.2f\nVb=%.2f\n",Va,Vb);
    	return 0;
    }
    

    Java :

    import java.util.*;
    public class Main {
    	public static void main(String[] arge)
    	{
    		float r,h,c,sa,sb,va,vb;
    		float PI=(float) 3.14;
    		Scanner scanner=new Scanner(System.in);
    	    r=scanner.nextFloat();
    	    h=scanner.nextFloat();
    	    java.text.DecimalFormat geshi =new java.text.DecimalFormat("#.00");
    	    c=PI*2*r;
    	    System.out.print("C1=");
    	    System.out.println(geshi.format(c));
    	    sa=r*r*PI;
    	    System.out.print("Sa=");
    	    System.out.println(geshi.format(sa));
    	    sb=4*PI*r*r;
    	    System.out.print("Sb=");
    	    System.out.println(geshi.format(sb));
    	    va=(float) (4.0/3*PI*r*r*r);
    	    System.out.print("Va=");
    	    System.out.println(geshi.format(va));
    	    vb=r*r*PI*h;
    	    System.out.print("Vb=");
    	    System.out.println(geshi.format(vb));
    	}
    }
    
    
    • 1

    C语言程序设计教程(第三版)课后习题4.8

    信息

    ID
    1490
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者