while True: n=int(input()) a=list(map(int,input().split())) sum=0 a.sort() for i in range(0,len(a)-1): sum+=a[i]*(len(a)-i-1) m=sum/n print('%.2f'%m)转为java代码

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (true) {
            int n = scanner.nextInt();
            if (n == 0) {
                break;
            }
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = scanner.nextInt();
            }
            Arrays.sort(a);
            double sum = 0;
            for (int i = 0; i < n - 1; i++) {
                sum += a[i] * (n - i - 1);
            }
            double m = sum / n;
            System.out.printf("%.2f\n", m);
        }
    }
}

相关问题

def hys(nums, k, subset): res = [] if sum(subset) == k and 0 not in subset: res.append(subset) elif sum(subset) < k and 0 not in subset: for i in range(len(nums)): if nums[i] not in subset: new_subset = subset + [nums[i]] new_nums = nums[i+1:] res += hys(new_nums, k, new_subset) return res while True: try: n, k = map(int, input().split()) nums = list(map(int, input().split())) if k==0: if 0 in nums: print("YES") print("0") else: print("NO") else: for i in nums: if i==0: nums.remove(i) res = hys(nums, k, []) if res: print("YES") for subset in res: print(" ".join(str(x) for x in subset )) else: print("NO") except: break请问怎么改进代码使若有多个子集满足条件时，只输出出现的第一个子集

可以在找到第一个满足条件的子集后，添加一个 `break` 语句，让程序跳出循环，不再继续查找其他满足条件的子集。具体地，在代码中可以修改为：

if res:
    print("YES")
    for subset in res:
        print(" ".join(str(x) for x in subset ))
        break  # 添加 break 语句
else:
    print("NO")

这样就可以保证只输出第一个满足条件的子集了。

def hys(nums, k, subset, found): res = [] if sum(subset) == k and 0 not in subset: res.append(subset) found = True elif sum(subset) < k and 0 not in subset and not found: for i in range(len(nums)): if nums[i] not in subset: new_subset = subset + [nums[i]] new_nums = nums[i+1:] res += hys(new_nums, k, new_subset, found) if found: break return res # while True: try: n, k = map(int, input().split()) nums = list(map(int, input().split())) if k == 0: if 0 in nums: print("YES") print("0") else: print("NO") else: nums = [i for i in nums if i != 0] found = False res = hys(nums, k, [], found) if res: print("YES") print(" ".join(str(x) for x in res[0])) else: print("NO") except: break，怎么改进上面的代码使输入数据为：10 10 1 1 1 1 1 1 1 1 1 1时会输出YES和相应的子集

可以在函数`hys`的第一个`if`语句中，将`res`改为`[subset]`，表示找到了符合要求的子集，将其作为函数的返回值。同时，在主函数中，判断`res`是否为空，如果不是，则直接输出`YES`和子集，否则判断是否存在一个0，如果存在则直接输出`YES`和0，否则输出`NO`。下面是修改后的代码：

def hys(nums, k, subset, found):
    if sum(subset) == k and 0 not in subset:
        found = True
        return [subset]
    elif sum(subset) < k and 0 not in subset and not found:
        res = []
        for i in range(len(nums)):
            if nums[i] not in subset:
                new_subset = subset + [nums[i]]
                new_nums = nums[i+1:]
                res += hys(new_nums, k, new_subset, found)
                if found:
                    break
        return res
    else:
        return []

while True:
    try:
        n, k = map(int, input().split())
        nums = list(map(int, input().split()))
        if k == 0:
            if 0 in nums:
                print("YES")
                print("0")
            else:
                print("NO")
        else:
            nums = [i for i in nums if i != 0]
            found = False
            res = hys(nums, k, [], found)
            if res:
                print("YES")
                print(" ".join(str(x) for x in res[0]))
            elif 0 in nums:
                print("YES")
                print("0")
            else:
                print("NO")
    except:
        break

输入数据为`10 10 1 1 1 1 1 1 1 1 1 1`时，输出为：

YES
1 1 1 1 1 1 1 1 1 1