突发奇想将自己的要求写成了伪代码形式，感觉程序猿相亲或者找游戏搭子（）是不是可以参考这种思路

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声明：

并不是真的找，只是感觉如果真的要送一张票找人陪我看（是的我确实有两张票（炫耀）），我会这样根据 XP 和实用性（？）来做决定。

叠甲：

代码水平不怎么样且玻璃心，如果觉得哪里有问题，希望能被指出问题而不是说我菜><说我菜我就哭（不是）
有些函数的具体内涵不一定和命名完全吻合（对我说的就是你leetcodeHard），因为不是真的找所以就写个名字，发出来主要是为了博看官一笑，希望不要做价值判断，非要做我也不介意但我大概率不会解释的

const PRIORITY = [
  {1, c => c.code.codeforces.rating >= 2000},
  {2, c => leetcodeHard(c)},
  {3, c => c.singing.lesMiserables.canSingAllTunes && c.isV2EXer},
  {4, c => c.videoGame.theWitness.hasFullAchievements},
  {5, c => c.body.hand.isPretty && willingHoldHands(c) && c.isV2EXer}
];

const CONDITIONS = [
  c => isStudent(c),
  c => c.gender == male,
  c => c.major == "math" || "CS" || "EE",
  c => knowsLesMiserables(c)
];

const finalJudging = [
  c => c.code.codeforces.rating,
  c => -c.education.schoolRank,
  c => -c.education.majorRank,
  c => c.singing.ability
]

Candidate select(candidates) {
  vector<vector> priorityCandidates(4);

  for (c : candidates) {
    int x = 0;
    for (p : PRIORITY) {
      if (p.second(c)) {
        x = p.first;
        break;
      }
    }
    if (x == 1) {
      return c;
      // 不会吧不会吧不会真有 cf2000 吧
    }else if (x > 0) {
      priorityCandidates[x-2].push_back(c);
    }
  }
  
  for (candidatesList : priorityCandidates) {
    Candidate result;
    int prev = 0;
    for (c : candidatesList) {
      int x = conditionsMetCount(c);
      if (x == 4) {
        return c;
      }
      if (x > prev) {
        prev = x;
        result = c;
      }
    }
    if (result) {
      return result;
    }
  }
    
  vector<Candidate> conditionCandidates;
  for (c : candidates) {
    bool meetAllConditions = true;
    for (con : conditions) {
      meetAllConditions &= con(c);
    }
    if (meetAllConditions) {
      conditionCandidates.push_back(c);
    }
  }
  
  if (conditionCandidates.empty()) {
    throw error("没有找到合适的人！");
  }
  
  sort(conditionCandidates, (a, b) => {
    for (f : finalJudging) {
      if (f(a) != f(b)) {
        return f(a) > f(b);
      }
    }
  });
  return conditionCandidates[0];
}

try {
  Candidate chosen = select(candidates);
  // 因为 candidates 数组本身就是按时间顺序排序的，所以有几个默认返回前者的情况。
} catch () {
  cry();
}

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