1 条题解

  • 0
    @ 2025-10-10 15:45:59

    C++ :

    #include<bits/stdc++.h>
    using namespace std;
    int main(){
    	int a[1500]={0},b[1500]={0},i,n;
    	double x=0,y=0;
    	double x1=0,x2=0;
    	cin>>n;
    	for(i=0;i<n;i++)cin>>a[i];
    	for(i=0;i<n;i++)cin>>b[i];
    	for(i=0;i<n;i++){
    		x1+=a[i];
    		x2+=b[i];
    	}
    	x2/=n;
    	x1/=n;
    	for(i=0;i<n;i++){
    		x+=(a[i]-x1)*(a[i]-x1);
    		y+=(b[i]-x2)*(b[i]-x2);
    	}
    	if(abs(x)<abs(y))cout<<"jia"<<endl;
    	else cout<<"yi"<<endl;
    	return 0;
    	
    }
    

    Java :

    import java.util.Scanner;
    
    public class Main {
    	public static void main(String[] args) {
    		Scanner sc = new Scanner(System.in);
    		int n = sc.nextInt();
    		int[] a =new int[n];
    		int[] b =new int[n];
    		for (int i =0;i<n;i++) {
    			a[i]=sc.nextInt();
    		}
    		for (int i =0;i<n;i++) {
    			b[i]=sc.nextInt();
    		}
    		int c = 0;
    		for (int i =0;i<n;i++) {
    			c=c+a[i];
    		}
    		double avea=(double)c/n;
    		int d = 0;
    		for (int i =0;i<n;i++) {
    			d=d+a[i];
    		}
    		double aveb=(double)d/n;
    		double x = 0;
    		for (int i =0;i<n;i++) {
    			x=x+(a[i]-avea)*(a[i]-avea);
    		}
    		double y = 0;
    		for (int i =0;i<n;i++) {
    			y=y+(b[i]-aveb)*(b[i]-aveb);
    		}
    		if (x<y) {
    			System.out.println("jia");
    		}
    		if (x>y) {
    			System.out.println("yi");
    		}
    	}
    }
    
    

    Python :

    n = int(input())
    a = input().split()
    b = input().split()
    av1 = 0
    av2 = 0
    su1 = 0
    su2 = 0
    s1 = 0
    s2 = 0
    l1 = []
    l2 = []
    for v in a:
        l1.append(int(v))
    for v in b:
        l2.append(int(v))
    for i in range(0, n):
        su1 += l1[i]
    av1 = su1 / n
    for i in range(0, n):
        su2 += l2[i]
    av2 = su2 / n
    for i in range(0, n):
        s1 += (l1[i] - av1) * (l1[i] - av1)
    for i in range(0, n):
        s2 += (l2[i] - av2) * (l2[i] - av2)
    if s1 < s2:
        print('jia')
    if s1 > s2:
        print('yi')
    
    • 1

    信息

    ID
    337
    时间
    1000ms
    内存
    256MiB
    难度
    10
    标签
    递交数
    3
    已通过
    2
    上传者