📜  求1000以下3或5的所有倍数之和python代码示例

📅  最后修改于: 2022-03-11 14:45:32.515000             🧑  作者: Mango

代码示例3
const findSum = n => {
  let countArr = []
  
  for(let i = 0; i <= n; i++) if(i % 3 === 0 || i % 5 === 0) countArr.push(i) 
  return countArr.reduce((acc , curr) => acc + curr)
}
console.log(findSum(5))