CS50 pset3：复数。 Print_winner function 在运行程序时无论输入如何都错误地打印多个候选

Question

CS50 的 pset3 复数要求您根据他们获得的票数打印“选举”的获胜者。 投票由用户输入。

在我必须打印获胜者之前，我的程序运行良好。 我已经尝试了几天尝试不同的方法，但无法理解它。

无论输入如何，我的代码当前都会输出多个候选者。 我猜变量“most_votes”和“election_winner”保留了代码顶部的分配值。

如果多个候选人获得相同数量的选票，如果他们获得最多票数和多个获胜者，我该如何重写它以打印单个获胜者？

谢谢你的帮助！

这是我的代码：

// Print the winner (or winners) of the election
void print_winner(void)
{
    int most_votes = candidates[0].votes;
    string election_winner = candidates[0].name;
    
    // Loop over each candidate 
    for (int i = 1; i < candidate_count; i++)
    {
        // Count how many votes each candidate received and determine who
        // received the most votes
        if (most_votes < candidates[i].votes)
        {
            most_votes = candidates[i].votes;
            election_winner = candidates[i].name;
        }
    }
    
    // Loop over candidates a second time
    for (int j = 1; j < candidate_count; j++)
    {
        // Count how many votes each candidate received
        // Compare number of votes of each candidate
        // Check if mutiple candidates have the most votes
        if (most_votes == candidates[j].votes && election_winner != candidates[j].name)
        {
            // Print all candidates with the most votes
            printf("%s\n", candidates[j].name);
        }
    }
    // Print the name of winner with most votes
    printf("%s\n", election_winner);
    return;
}

Answer 1

如果可以有多个获胜者，您不需要（也不应该拥有）变量election_winner 。

1. 在第一个循环中只确定most_votes的值。
2. 第二次循环候选者（从 0 开始，而不是像您当前所做的那样从 1 开始）。 如果候选人的票数等于most_votes ，那么您就知道候选人是赢家，您可以将姓名打印出来，或者其他任何需要的内容。

Answer 2

感谢您的帮助，我能够弄清楚。 将其简化为@vmt 建议并有效！

将变量“most_votes”更改为“0”而不是“candidates[0].votes”可以更容易理解。

// Print the winner (or winners) of the election
void print_winner(void)
{
    int most_votes = 0;
    
    // Loop over each candidate 
    for (int i = 0; i < candidate_count; i++)
    {
        // Count which candidate has the most votes and assign them most_votes
        if (most_votes < candidates[i].votes)
        {
            most_votes = candidates[i].votes;
        }
    }
    
    // Loop again over candidates
    for (int j = 0; j < candidate_count; j++)
    {
        // Check if mutiple candidates have the most votes
        if (most_votes == candidates[j].votes)
        {
            // Print all candidates with the most votes
            printf("%s\n", candidates[j].name);
        }
    }
    return;
}