1 条题解

  • 0
    @ 2025-4-7 21:28:47

    C :

    #include<stdio.h>
    int main()
    {
    	int i,m,n;
    	double a[55],b[55],sum1,sum2;
    	scanf("%d%d",&n,&m);
    	for(sum1=0,i=0;i<n;i++)
    	{
    		scanf("%lf",&a[i]);
    		sum1+=a[i];
    	}
    	for(sum2=0,i=0;i<m;i++)
    	{
    		scanf("%lf",&b[i]);
    		sum2+=b[i];
    	}
    	printf("%.2lf %.2lf\n",sum1/n,sum2/m);
    	return 0;
    }
    

    C++ :

    #include <stdio.h>
    int main() {
    	float average(float array[], int n);
    	float score[50];
    	float aver1, aver2;
    	int n, m, i;
    	scanf("%d%d", &n, &m);
    	for (i = 0;i < n;i++)
    		scanf("%f", &score[i]);
    	aver1 = average(score, n);
    	for (i = 0;i < m;i++)
    		scanf("%f", &score[i]);
    	aver2 = average(score, m);
    	printf("%.2f %.2f\n", aver1, aver2);
    	return 0;
    }
    float average(float array[], int n) {
    	int i;
    	float aver, sum = array[0];
    	for(i = 1;i < n;i++)
    		sum = sum + array[i];
    	aver = sum / n;
    	return aver;
    }
    
    

    Pascal :

    var
      i,n,m:longint;
      k,s,d:double;
    begin
      readln(n,m);
      for i:=1 to n do
      begin
        read(k);
        s:=s+k;
      end;
      for i:=1 to m do
      begin
        read(k);
        d:=d+k;
      end;
      writeln(s/n:0:2,' ',d/m:0:2);
    end.
    
    
    

    Java :

    import java.util.*;
    
    public class Main {
    	public static void main(String[] args) {
    		Scanner in = new Scanner(System.in);
    		int n=in.nextInt();
    		int m=in.nextInt();
    		double a=0,b=0;
    		for(int i=1;i<=n;i++)
    			a+=in.nextDouble();
    		for(int j=1;j<=m;j++)
    			b+=in.nextDouble();
    		System.out.printf ("%.2f %.2f\n",a/n,b/m);
    	}	
    }
    
    

    Python :

    raw_input()
    a = [float(i) for i in raw_input().split()]
    a = sum(a)/len(a)
    b = [float(i) for i in raw_input().split()]
    b = sum(b)/len(b)
    print '%.2f %.2f' % (a,b)
    
    • 1

    信息

    ID
    1316
    时间
    1000ms
    内存
    32MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者