def find_max_min(nums): max_num = max(nums) min_num = min(nums) return max_num, min_num nums = [int(x) for x in input("请输入一组整数,以空格分隔:").split()] max_num, min_num = find_max_min(nums) print("最大值:", max_num) print("最小值:", min_num) ```反馈...
n,m = map(int,input().split()) nums = [] for i in range(m): nums.append(int(input())) class sol: def __init__(self): # self.seen = set() def dfs(self,nums,n,m,pos,step): # 停止条件:如果是最后一次看时间后走的 if step == m: if pos not in self.seen: self.seen...
nums = list(map(int, input().split())) nums[:] = sorted(nums) list1 = [] l = 0 r = len(nums)-1 mins = nums[l]+nums[r] while rgt;l: if abs(nums[l]_牛客网_牛客在手,offer不愁
nums1=list(map(int,input.split)) nums2=list(map(int,input.split)) m=len(nums1) n=m//3 dic=dict foriinrange(n): forjinrange(3): dic[nums2[i*3+j]]=i nums=[dic[num]fornuminnums1] 检查某个组是否已经排好队 在后面的模拟过程中,我们就可以通过构建一个函数check_group(nums, idx...
dp[i][1] = max(dp[i-1][0]+x, dp[i-1][1]+nums[i]) # 更新答案 ans = max(ans, max(dp[i])) returnans t = int(input) # 遍历t次,表示t次询问 for_inrange(t): # 获得本次询问的数组长度n和可以修改数字x n, x = map(int, input.split) ...
there is no nontrivial sense of variable-length encoding that applies generically while still being interesting.2 This is easiest to see by example (Illustrated in Figure 1 and revisited as Example 2 in This material was presented in part at the IEEE Int Symp Inform Theory, Adelaide, Australia...
n, k = map(int, input().split()) nums = list(map(int, input().split())) nums.sort() left, right = n-2, n - 1 for _ in range(min(k, n-1)): nums[right] = nums[left] * nums[right] % (10**9+7) nums[left] = 1 left -= 1 print(sum(nums) % (10**9+7)) ...
n = int(input()) nums = list(map(int,input().split())) nums = sorted(nums) first = [] second = [] for i in range(n//2): first.append(nums[i]) for i in range(n//2,n): second.append(nums[i]) if n % 2: temp = second[1] else: temp = -first[-1] x1 = sum(...
public int findMaxConsecutiveOnes(int[] nums) { int max = 0, cur = 0; for (int x : nums) { cur = x == 0 ? 0 : cur + 1; max = Math.max(max, cur); } return max;} 一个数组元素在 [1, n] 之间,其中一个数被替换为另一个数,找出重复的数和丢失的数 ...
nums2=list(map(int,input.split)) #m个同学 m=len(nums1) #n个组 n=m//3 #构建映射哈希表 dic=dict #对于nums2,每三个元素映射到组编号i foriinrange(n): forjinrange(3): dic[nums2[i*3+j]]=i #根据映射哈希表,构建数组nums