CS50 Tideman“当一些配对并列时打印选举获胜者”失败

Question

编辑： 固定的！必须为(winscore winscores[11];) 设置一个值，其余代码都很好，但可能更简洁一些。最后的条件是：

for (int i = 0; i < candidate_count; i++)
    {
        if (winscores[10].score >= winscores[i].score)
        {
            printf("%s\n", candidates[winscores[i].candidate]);        
        } 
    }

现在所有检查都通过了！

---

我正在开发 CS50 Tideman，但我的代码没有通过最终检查：“print_winner 在某些配对打成平手时打印选举获胜者”。

获胜者应该在锁定矩阵中拥有最多的false，所以我计算false的最大数量，然后按false的数量降序排列。然后我打印出false 数量最多的候选人。然后我检查是否有更多候选人具有相同数量的false。如果有的话，我也打印出来。如果多个候选人拥有相同数量的false，则应该打印获胜者或获胜者......至少我认为，但它仍然没有通过最终检查。我不确定我的代码有什么问题，因此我们将不胜感激！

typedef struct {
  int candidate;
  int score;
} winscore;

winscore winscores[11];

// Print the winner of the election
void print_winner(void) {
  // int maxscore = 0;
  // int maxscore[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};

  for (int i = 0; i < candidate_count; i++) {
    for (int j = 0; j < candidate_count; j++) {
      if (locked[i][j] == false) {
        winscores[i].candidate = i;
        winscores[i].score++;
      }
    }
  }

  for (int i = 0; i < candidate_count; i++) {
    for (int j = 0; j < candidate_count; j++) {
      if (winscores[i].score < winscores[j].score) {
        winscores[10].candidate = winscores[i].candidate;
        winscores[10].score = winscores[i].score;

        winscores[i].candidate = winscores[j].candidate;
        winscores[i].score = winscores[j].score;

        winscores[j].candidate = winscores[10].candidate;
        winscores[j].score = winscores[10].score;
      }
    }
  }

  printf("%s\n", candidates[winscores[0].candidate]);

  for (int i = 0; i < 10; i++) {
    if (winscores[0].score == winscores[i].score) {
      printf("%s\n", candidates[winscores[i].candidate]);
    }
  }

  return;
}

Answer 1

至少有这个问题：winscores[i] 缺少初始化导致winscores[i].score++; 出现问题。

// winscore winscores[11];
winscore winscores[11] = { 0 };