Complete the function kangaroo in the editor below.
kangaroo has the following parameter(s):
int x1, int v1: starting position and jump distance for kangaroo 1
int x2, int v2: starting position and jump distance for kangaroo 2
在下面的編輯器中完成袋鼠功能。 袋鼠有以下參數: int x1, int v1: 袋鼠1的起始位置和跳躍距離 int x2、int v2:袋鼠2的起始位置和跳躍距離
Example 1:
x1 = 2 v1 = 1 x2 = 1 v2 = 2 After one jump, they are both at x = 3, (x1 + v1 = 2 + 1, x2 + v2 = 1 + 2), so the answer is YES.
solution:
首先我們先了解 x1, v1, x2, v2 分別代表兩隻袋鼠的起始位置與速度。
接著我們計算兩隻袋鼠的相遇時間(meet = (x2 – x1) / (v1 – v2)),
並回傳是否相遇。
Code 1: BigO(1)
'use strict'; const fs = require('fs'); process.stdin.resume(); process.stdin.setEncoding('utf-8'); let inputString = ''; let currentLine = 0; process.stdin.on('data', function(inputStdin) { inputString += inputStdin; }); process.stdin.on('end', function() { inputString = inputString.split('\n'); main(); }); function readLine() { return inputString[currentLine++]; } /* * Complete the 'kangaroo' function below. * * The function is expected to return a STRING. * The function accepts following parameters: * 1. INTEGER x1 * 2. INTEGER v1 * 3. INTEGER x2 * 4. INTEGER v2 */ function kangaroo(x1, v1, x2, v2) { // Write your code here let meet = (x2 - x1) / (v1 - v2); if (meet === Math.round(meet) && meet > 0 && meet !== Infinity) { return "YES"; } return 'NO'; } function main() { const ws = fs.createWriteStream(process.env.OUTPUT_PATH); const firstMultipleInput = readLine().replace(/\s+$/g, '').split(' '); const x1 = parseInt(firstMultipleInput[0], 10); const v1 = parseInt(firstMultipleInput[1], 10); const x2 = parseInt(firstMultipleInput[2], 10); const v2 = parseInt(firstMultipleInput[3], 10); const result = kangaroo(x1, v1, x2, v2); ws.write(result + '\n'); ws.end(); }
FlowChart:
Example 1
x1 = 2 v1 = 1 x2 = 1 v2 = 2 meet = -1 / -1 //1 meet === Math.round(meet) //meet = 1 === Math.round(meet) meet > 0 //meet = 1 > 0 meet !== Infinity //不等於無窮大 return "YES";