PAT-BASIC1011——A+B 和 C
阿新 • • 發佈:2018-12-16
題目描述:
知識點:資料越界
思路:用long型變數儲存輸入值,以防越界
時間複雜度是O(n),其中n為輸入的資料對數。空間複雜度是O(1)。
C++程式碼:
#include<iostream> using namespace std; int main() { int count; cin >> count; long num1 = 0; long num2 = 0; long num3 = 0; for (int i = 1; i <= count; i++) { cin >> num1 >> num2 >> num3; if (num1 + num2 > num3) { cout << "Case #" << i << ": " << "true" << endl; } else { cout << "Case #" << i << ": " << "false" << endl; } } }
C++解題報告:
JAVA程式碼:
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int count = Integer.parseInt(scanner.nextLine()); for (int i = 1; i <= count; i++) { String[] strings = scanner.nextLine().split("\\s+"); Long[] nums = new Long[3]; for(int j = 0; j < 3; j++){ nums[j] = Long.parseLong(strings[j]); } if(nums[0] + nums[1] > nums[2]){ System.out.print("Case #" + i + ": " + true); }else{ System.out.print("Case #" + i + ": " + false); } if(i != count){ System.out.println(); } } } }
JAVA解題報告: