억억단을 외우자

Link

https://school.programmers.co.kr/learn/courses/30/lessons/138475

Deadline

Feb 14, 2023

Status

Archived

Type

풀이

재영


const solution = (e, starts) => {
  const DP = new Array(e + 1).fill(0);

  const results = new Array(e + 1).fill(1);

  for (let num = 2; num <= e; num += 1) {
    for (let divider = num; divider <= e; divider += num) {
      DP[divider] += 1;
    }
  }

  for (let i = results.length - 1; i >= 0; i -= 1) {
    if (i === results.length - 1) {
      results[i] = i;
      continue;
    }

    const prevMaxIdx = results[i + 1];

    results[i] = DP[i] >= DP[prevMaxIdx] ? i : prevMaxIdx;
  }

  return starts.map((start) => results[start]);
};

효성


function solution(e, starts) {
    const divisorArr = new Array(e + 1).fill(0);
    for (let i = 1; i <= e; i += 1) {
        for (let j = i; j <= e; j += i) {
            divisorArr[j]++;
        }
    }
    
    const maxCountArr = new Array(e + 1).fill(0);
    
    let curDivisor = 0;
    for (let i = e; i >= 1; i -= 1) {
        if (curDivisor <= divisorArr[i]) {
            curDivisor = divisorArr[i];
            maxCountArr[i] = i;
        } else {
            maxCountArr[i] = maxCountArr[i + 1];
        }
    }
    
    return starts.map((v) => maxCountArr[v]);
}

은찬

오랜만에 왔습니다ㅎㅎ

오래걸렸네요…


function solution(e, starts) {
    const answer = [];
    const arr = Array.from({length: 5000001}, () => 0);
    const max = Array.from({length: 5000001}, () => 0);
    let count = 0;
    let index = e;
    
    for(let i = 1; i <= e; i++) {
        for(let j = i; j <= e; j += i) {
            arr[j]++;
        }
    }
    
    for(let i = e; i >= min; i--){
        if(i == e){
            max[i] = i;
            count = arr[i];
        } else{
            if(arr[i] < count){
                max[i] = index;
            } else {
                max[i] = i;
                count = arr[i];
                index = i;
            }
        }
    }
    
    for(let i = 0; i < starts.length; i++) {
        const start = starts[i];
        
        answer.push(max[start]);
    }
    
    return answer;
}