未能正确计数正确添加_PAIRS TIDEMAN CS50

Question

有人可以帮助我了解为什么我的代码失败了以下要点吗？此功能旨在通过获胜对更新对阵列，并相应地增加pair_count变量。我的函数代码如下：

void add_pairs(void)
{
    float cc = candidate_count;// have to set to float to make sure a float number is given as the result
    float majority = (cc / 2);
    // loop through preferences array using nested loop, find preferences for candidates, assign winner or loser depending on scores, increment pair count for each
    // loop nested for, conditional if preferences[i][j] > candidate_count / 2, winner = i loser =j, pair_count += 1. Add pair to pairs array.
    for (int i = 0; i < candidate_count; i++)// loops through preferences array
    {
        for (int j = 0; j < candidate_count; j++)
        {
            if (preferences[i][j] > (majority)) // increments pair count in the event there is a majority for one candidate over another
            {
                pair_count += 1;
                for (int k = 0; k < pair_count; k++) // adds that pair to the pairs array
                {
                    pairs[k].winner = i;
                    pairs[k].loser = j;
                }
            }
        }
    }
    printf("pc:%i\n", pair_count);
    return;
}

拟合较大的程序：

#include <cs50.h>
#include <stdio.h>
#include <string.h>

// Max number of candidates
#define MAX 9

// preferences[i][j] is number of voters who prefer i over j
int preferences[MAX][MAX];// if x prefer then the margin is x/voters - x

// locked[i][j] means i is locked in over j - adjacency matrix
bool locked[MAX][MAX];

// Each pair has a winner, loser
typedef struct
{
    int winner;
    int loser;
}
pair;

// Array of candidates
string candidates[MAX];
pair pairs[MAX * (MAX - 1) / 2];

int pair_count;
int candidate_count;

// Function prototypes
bool vote(int rank, string name, int ranks[]);
void record_preferences(int ranks[]);
void add_pairs(void);
void sort_pairs(void);
void lock_pairs(void);
void print_winner(void);

int main(int argc, string argv[])
{
    // Check for invalid usage
    if (argc < 2)
    {
        printf("Usage: tideman [candidate ...]\n");
        return 1;
    }

    // Populate array of candidates
    candidate_count = argc - 1;
    if (candidate_count > MAX)
    {
        printf("Maximum number of candidates is %i\n", MAX);
        return 2;
    }
    for (int i = 0; i < candidate_count; i++)
    {
        candidates[i] = argv[i + 1];
    }

    // Clear graph of locked in pairs
    for (int i = 0; i < candidate_count; i++)
    {
        for (int j = 0; j < candidate_count; j++)
        {
            locked[i][j] = false;
        }
    }

    pair_count = 0;
    int voter_count = get_int("Number of voters: ");

    // Query for votes
    for (int i = 0; i < voter_count; i++)
    {
        // ranks[i] is voter's ith preference
        int ranks[candidate_count];

        // Query for each rank
        for (int j = 0; j < candidate_count; j++)
        {
            string name = get_string("Rank %i: ", j + 1);

            if (!vote(j, name, ranks))
            {
                printf("Invalid vote.\n");
                return 3;
            }
        }

        record_preferences(ranks);

        printf("\n");
    }

    add_pairs();
    sort_pairs();
    lock_pairs();
    print_winner();
    return 0;
}

以下是与上面的add_pairs代码相对应的check50结果

:( add_pairs在没有纽带时会生成正确的对计数

add_pairs function did not produce 3 pairs

:( add_pairs在存在纽带时会生成正确的对时计数

add_pairs function did not produce 2 pairs

:)带有获胜对的填充对阵列

:) add_pairs

如果我以./tideman asd为./tideman asd，则在以下偏好a，s，s，d -s，a，a，d -a，a，s，s，s， d。我在函数末尾的printf语句显示了一个正确的paim_count_ 3，但我仍然按照上述点失败