D24_ ALG101-Unit 5

LIOJ 1029：加減乘除

Output a '+', '-', '*', '/' b，the OJ input would be 'a + b' or 'a * b'

var readline = require('readline');

var lines = []
var rl = readline.createInterface({
  input: process.stdin
});

rl.on('line', function (line) {
  lines.push(line)
});

rl.on('close', function() {
  solve(lines)
})

function solve(lines) {
    let input = lines[0].split(' ')
    let a = Number(input[0])
    let symbol = input[1]
    let b = Number(input[2])

    if (symbol === '+') {
        console.log(a + b)
    } else if (symbol === '-') {
        console.log(a - b)
    } else if (symbol === '*') {
        console.log(a * b)
    } else {
        console.log(a / b)
    }
}

LIOJ 1030：判斷迴文

Find if a reversed string equals to the original string

在解這題的時候，我 i 的初始值設錯了，應該是要 input.length - 1，我忘了減一，所以一直都是 False; I forgot to setup the string length with 'input.length - 1', so the answer's always 'False'

var readline = require('readline');

var lines = []
var rl = readline.createInterface({
  input: process.stdin
});

rl.on('line', function (line) {
  lines.push(line)
});

rl.on('close', function() {
  solve(lines)
})

function solve(lines) {
    let input = lines[0]
  if (reverse(input) === input) {
        console.log('True')
    } else {
        console.log('False')
    }
}

function reverse(input) {
    let result = ''
    for(i = input.length - 1; i>=0; i--) {
        result += input[i]
    }
    return result
}

LIOJ 1031：完全平方和

Find the summation of perfect square within a certain number, ex 30: 1 + 4 + 9 + 16 + 25 = 55 The answer should be 55

卡在要如何判斷一個數是否為整數，後來想想，只要把從 1 開始把每個數平方，從 1 到小於題目指定所有的數，全部加起來就是答案了。
Maybe I don't need to judge if a number is an integer, just find the square numbers smaller than a certain number and then make a summation.

var readline = require('readline');

var lines = []
var rl = readline.createInterface({
  input: process.stdin
});

rl.on('line', function (line) {
  lines.push(line)
});

rl.on('close', function() {
  solve(lines)
})

function solve(lines) {
    let n = Number(lines[0])
    let ans = 0
    for(i = 1; i <= n; i++) {
        if(i*i <= n) {
            ans += i * i
        }
    }
    console.log(ans)
}

LIOJ 1032：平面距離計算

有三個問題：
1. 怎麼取到資料？我卡在這個地方，要注意的是 lines[0] 其實就是讓我知道計算的次數，另外 x1, x2, y1, y2 那四個數不可以寫死，所以要找到每一組的開頭，題目的這句話是重點「每一筆測試資料會有 4 個整數」
2. 寫第一個函式計算整個過程
3. 怎麼取到小數點第二位 result.toFixed(2)
解題過程：我真的是太粗心了，要不就忘記轉成數字，要不就 x1, x2 順序寫錯
要看清楚題目

var readline = require('readline');

var lines = []
var rl = readline.createInterface({
  input: process.stdin
});

rl.on('line', function (line) {
  lines.push(line)
});

rl.on('close', function() {
  solve(lines)
})

function solve(lines) {
    let times = Number(lines[0])
  for(let i = 1; i <= times; i++) {
    //i 就是筆數
        //我覺得這題最難的是理解這一段，從 lines[5] 往回推 lines[1]
        let start = (i - 1) * 4 + 1 
        let x1 = Number(lines[start])
        let y1 = Number(lines[start + 1])
        let x2 = Number(lines[start + 2])
        let y2 = Number(lines[start + 3])
        console.log(distance(x1, y1, x2, y2))
  }
}

function distance(x1, y1, x2, y2) {
    let result = Math.sqrt(
        Math.abs(x1 - x2) * Math.abs(x1 - x2) + 
        Math.abs(y1 - y2) * Math.abs(y1 - y2)
    )   
    return result.toFixed(2)
}