题解 | #数组分组#

题解如下，请参考：

const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;

void async function () {
    // Write your code here
    let num = Number(await readline())
    let arr = (await readline()).split(' ').map(Number)
    //构建5的倍数、3的倍数的数组arr5和arr3，及剩余数字组成的数组arr
    let arr5 = [], arr3 = []
    for(let i=0; i<arr.length; i++){
        if(arr[i]%5 == 0){
            arr5.push(arr[i])
            arr.splice(i, 1)
            i--
        }else if(arr[i]%3 == 0){
            arr3.push(arr[i])
            arr.splice(i, 1)
            i--
        }
    }
  
    //将arr5和arr3的所有数值加和求出sum5和sum3
    let sum5=0, sum3=0
    if(arr5.length>0){
        for(let k of arr5){
            sum5 += k
        }
    }
    if(arr3.length>0){
        for(let k of arr3){
            sum3 += k
        }
    }

    //将sum5和sum3的差值和arr数组中的值放在同一数组中，并将将每个数字转化成正整数
    let newArr = [sum3-sum5, ...arr].map((n) => Math.abs(n))
    newArr.sort((a,b) => b-a)
    while(newArr.length>1){
        let big = newArr.shift()
        let small = newArr.shift()
        newArr.push(big-small)
        newArr.sort((a,b) => b-a)
    }
    console.log(newArr[0]==0)
}()