1 条题解

  • 0
    @ 2023-6-11 12:20:58

    C :

    #include<stdio.h>
    int main()
    {
    	int x,y,z;
    	scanf("%d%d%d",&x,&y,&z);
    	int a[3];
    	a[0]=x;
    	a[1]=y;
    	a[2]=z;
    	int t,i,j;
    	for(i=0;i<2;i++){
    		for(j=i+1;j<3;j++){
    			if(a[i]>a[j]){
    				t=a[i];
    				a[i]=a[j];
    				a[j]=t;
    			} 
    		} 
    	}
    	int n=a[0]+a[2];
    	printf("%d\n%.1f",n,n/2.0);
    	return 0;
    }
    
    

    C++ :

    #include <bits/stdc++.h>
    using namespace std;
    
    int main() {
    	int ma,mi,a[3];
    	cin>>a[0]>>a[1]>>a[2];
    	sort(a,a+3);
    	
    	cout<<a[0]+a[2]<<endl;
    	cout<<fixed<<setprecision(1)<<(a[0]+a[2]) / 2.0;
    }
    
    

    Java :

    
    import java.util.Scanner;
    
    public class Main {
    
    	public static void main(String[] args) {
    		Scanner sc = new Scanner(System.in);
    		int x = sc.nextInt();
    		int y = sc.nextInt();
    		int z = sc.nextInt();
    		
    		int n;
    		double m;
    		
    		if(x>=y && x>=z && z<=y) {
    			n = x + z;
    			m = (x + z)*1.0/2;
    			System.out.println(n);
    			System.out.println(m);
    		}
    		
    		if(x>=y && x>=z && y<=z) {
    			n = x + y;
    			m = (x + y)*1.0/2;
    			System.out.println(n);
    			System.out.println(m);
    		}
    		
    		
    		if(y>=x && y>=z && z<=x) {
    			n = y + z;
    			m = (z + y)*1.0/2;
    			System.out.println(n);
    			System.out.println(m);
    		}
    		
    		if(y>=x && y>=z && x<=z) {
    			n = x + y;
    			m = (x + y)*1.0/2;
    			System.out.println(n);
    			System.out.println(m);
    		}
    		
    		if(z>=y && z>=x && x<=y) {
    			n = x + z;
    			m = (x + z)*1.0/2;
    			System.out.println(n);
    			System.out.println(m);
    		}
    		
    		if(z>=y && z>=x && y<=x) {
    			n = y + z;
    			m = (y + z)*1.0/2;
    			System.out.println(n);
    			System.out.println(m);
    		}
    
    		sc.close();
    	}
    
    }
    
    
    • 1

    信息

    ID
    2658
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者