publicstaticvoidmain(String[] args){Scannerscanner=newScanner(System.in);intn=scanner.nextInt();int[] subSum =newint[n]; subSum[0] = scanner.nextInt();for(inti=1;i<n;i++) { subSum[i] = subSum[i-1]+scanner.nextInt(); }intresult=0;for(inti=0;i<n;i++) {if(subSum[i]...
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sumOfCutCounts += (length/i - 1); //总长被定长整除,可以少切一次 } else { sumOfCutCounts += (length/i); } sumOfCutWastes += (length%i); //统计浪费的零头 } int profit = sumOfLengths*metal_price - sumOfCutCounts*cost_per_cut - sumOfCutWastes*metal_price; //总价 - 切割费用 ...
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423.1415WelcometoHackerRank's Java tutorials! Sample Output String:WelcometoHackerRank's Java tutorials!Double:3.1415Int:42 Hi, I don't understand why we have todo a sc.nextLine(), then do it again sc.nextLine()... Sometimes you have toclear the bufferto print the strings by command sc.ne...
#TitleSolutionTimeSpaceDifficultyPointsNote Solve Me First Java C# O(1) O(1) Easy 1 Simple Array Sum Java C# O(n) O(1) Easy 10 Compare the Triplets Java C# O(1) O(1) Easy 10 A Very Big Sum Java C# O(n) O(1) Easy 10 Diagonal Difference Java C# O(n) O(1)...
function printFirstItem(arrayOfItems) { console.log(arrayOfItems[0]); }This function runs in O(n) time (or "linear time"), where n is the number of items in the array. If the array has 10 items, we have to print 10 times. If it has 1,000 items, we have to print 1,000 ...
#TitleSolutionTimeSpaceDifficultyPointsNote Print the Elements of a Linked List C++ O(n) O(1) Easy 5 Reverse a Linked List Java O(n) O(1) Easy 5 Compare Two Linked Lists C++ O(n) O(1) Easy 5 Delete a node C++ O(n) O(1) Easy 5 ...
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