Python3 解法, 执行用时: 428ms, 内存消耗: 4712K, 提交时间: 2023-08-05 08:35:41

n, v, u = map(float, input().split())
c = list(map(float, input().split()))
d = list(map(float, input().split()))
t = 0
for i in range(int(n)):
    for j in range(int(n)):
        t += n * u / (c[i] - j * d[i] - v)
every = t / n
print("%.3f" % every)

C++(clang++ 11.0.1) 解法, 执行用时: 6ms, 内存消耗: 396K, 提交时间: 2022-08-06 23:14:36

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e3+5;
double c[N],d[N],f[N];
int main(){
    int n;
    double v,u;
    cin>>n>>v>>u;
    double ans=0;
    for(int i=1;i<=n;i++)
        cin>>c[i];
    for(int i=1;i<=n;i++) cin>>d[i];
    for(int i=1;i<=n;i++){
        for(int j=0;j<n;j++){
            f[i]+=1.0/(c[i]-v-j*d[i]);
        }
        ans+=f[i]*u;
    }
    printf("%.3lf\n",ans);
    return 0;
}

Java 解法, 执行用时: 337ms, 内存消耗: 20264K, 提交时间: 2023-08-08 15:56:49

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n=in.nextInt();
		double v=in.nextDouble();
		double u=in.nextDouble();
		double[] c=new double[n];
		for(int i=0;i<n;i++) {
			c[i]=in.nextDouble();
		}
		double sum=0.0;
		for(int i=0;i<n;i++) {
			double d=in.nextDouble();
			for(int j=1;j<=n;j++) {
				sum+=u/(c[i]-v-(j-1)*d);
			}
		}
		System.out.printf("%.3f",sum);
	}

}