c语言：输入两个整数,计算并输出这两个整数的和·平均数·最大值·最小值？

Posted 2023-04-15

你好，程序如下

#include<stdio.h>
void main()

float a,b,sum,avg,min,max;
printf("请输入两个整数:\n");
scanf("%f%f",&a,&b);
sum=a+b;
avg=(a+b)/2;
if(a>=b)

max=a;
min=b;

else
max=b;
min=a;

printf("这两个整数的和是: %f\n",sum);
printf("这两个整数的平均数是: %f\n",avg);
printf("这两个整数中较小的数是: %f\n",min);
printf("这两个整数中较大的数是: %f\n",max);


有疑问提出
望采纳 参考技术A #include<stdio.h>
main()

int a,b;
int num,max,min;
float ave;
printf("input two numbers:\n");
scanf("%d %d",&a,&b);
num = a+b;
ave = (float) (a+b) /2.0 ;
if(a>b) max=a;min=b;
elsemax=b;min=a;

printf("最大:%d\n最小:%d\n和:%d\n平均:%f\n",max,min,num,ave);

system("pause");

求JAVA的两整数输出最大最小值以及两个数的和差积商的程序

急急急！！！！跪求高手给答案．．．感激不尽．．．

参考技术A /**
* @(#)Zhidao1.java
*
*
* @author 智鸱儿—zhichier
* @version 1.00 己丑年二月廿六日-2009/3/22
*/

public class Zhidao1

private int x;
private int y;

//构造方法一—默认构造方法
public Zhidao1()
x = 1;
y = 1;


//构造方法二—带参数的构造方法
public Zhidao1(int x, int y)
this.x = x;
this.y = y;


//求和函数
public int qiuHe()
return x + y;


//求差函数
public int qiuCha()
return x - y;


//求积函数
public int qiuJi()
return x * y;


//求商函数
public double qiuShang()
return y == 0 ? 0.0 : ((double)x / (double)y);


//求余函数
public int qiuYu()
return y == 0 ? 0 : x % y;


//求最大值函数
public int qiuZuida()
return x > y ? x : y;


//求最小值函数
public int qiuZuixiao()
return x < y ? x : y;


public static void main(String args[])

//用默认构造方法，x和y被分别设置为1
Zhidao1 zd1= new Zhidao1();

System.out.println("x和y的和是：" + zd1.qiuHe());
System.out.println("x和y的差是：" + zd1.qiuCha());
System.out.println("x和y的积是：" + zd1.qiuJi());
System.out.println("x和y的商是：" + zd1.qiuShang());
System.out.println("x和y的余是：" + zd1.qiuYu());
System.out.println("x和y的最大值是：" + zd1.qiuZuida());
System.out.println("x和y的最小值是：" + zd1.qiuZuixiao());

//用带参数构造方法，x和y被分别设置为3和5
Zhidao1 zd2 = new Zhidao1(3, 5);

System.out.println("x和y的和是：" + zd2.qiuHe());
System.out.println("x和y的差是：" + zd2.qiuCha());
System.out.println("x和y的积是：" + zd2.qiuJi());
System.out.println("x和y的商是：" + zd2.qiuShang());
System.out.println("x和y的余是：" + zd2.qiuYu());
System.out.println("x和y的最大值是：" + zd2.qiuZuida());
System.out.println("x和y的最小值是：" + zd2.qiuZuixiao());


参考技术B import java.io.*;

class SEV
public static void main(String[] args) throws IOException
int max,min;
double sh;
InputStreamReader reader = new InputStreamReader(System.in);
BufferedReader input = new BufferedReader(reader);
System.out.print("please input m:");
String r = input.readLine();
int m = Integer.parseInt(r);
System.out.print("please input n:");
String s = input.readLine();
int n = Integer.parseInt(s);
if(m>n) max=m;min=n;
else min=m;max=n;
sh=(double)max/min;
System.out.println("max="+max);
System.out.println("min="+min);
System.out.println("max+min="+(max+min));
System.out.println("max-min="+(max-min));
System.out.println("max*min="+(max*min));
System.out.println("max/min="+sh);

本回答被提问者采纳 参考技术C 以前也有过类似的经历 参考技术D 这。。。太简单了 作业题吧 第5个回答  2009-09-21 HOHO~我的简单！

public class C

public static void main(String[] args)
int a,b;
System.out.println("请依次输入2个整数！");
Scanner scan1=new Scanner(System.in);
int i=scan1.nextInt();
Scanner scan2=new Scanner(System.in);
int j=scan2.nextInt();
a=i>j?i:j;
b=i<j?i:j;
System.out.println("最大数为:"+a);
System.out.println("最小数为:"+b);
double c=(double)i/(double)j;
System.out.println("和："+(i+j)+"商："+c+"积："+(i*j)+"差："+(i-j));