如何从一个数组中找到每个分支的百分比？在C

Question

（问题底部的编辑。）

例如，让我们有一个数组，其中填充了一年中每个月的一些元素。用户首先输入他想输入数据的分支数量，因此程序会自行循环并将其存储在one数组中，我们制作了另一个函数，并让其计算了输入的所有数据的总和用户。现在，该怎么做，我们需要使其显示每个分支整体的百分比。

Input: Please enter the number of branches you'd like to input data about : 2
Branch 1
Enter the data for the month 1 : 123
Enter the data for the month 2 : 123
Enter the data for the month 3 : 123
Enter the data for the month 4 : 123
Enter the data for the month 5 : 123
Enter the data for the month 6 : 123
Enter the data for the month 7 : 123
Enter the data for the month 8 : 123
Enter the data for the month 9 : 123
Enter the data for the month 10 : 123
Enter the data for the month 11 : 123
Enter the data for the month 12 : 123

Branch 2
Enter the data for the month 1 : 123
Enter the data for the month 2 : 123
Enter the data for the month 3 : 123
Enter the data for the month 4 : 123
Enter the data for the month 5 : 123
Enter the data for the month 6 : 123
Enter the data for the month 7 : 123
Enter the data for the month 8 : 123
Enter the data for the month 9 : 123
Enter the data for the month 10 : 123
Enter the data for the month 11 : 123
Enter the data for the month 12 : 123


The sum of all the branches income combined is = 2952.000000 

Required output: Branch 1 has : 50%
 Branch 2 has : 50%

（是的，这是我第一次在这里问问题）

（（如果需要编辑，这里是完整的代码2：感谢Armali修复了旧的破烂代码btw）

#include <stdio.h>
#include <stdlib.h>

void Data_Of_Branches(int number_of_branches, float arr_brms[number_of_branches][12])
{
  int i;
  int j;
  for (i = 0; i < number_of_branches; i++)
  {
    printf("Branch %d\n", i + 1);
    for (j = 0; j < 12; j++)
    {
      printf("Enter the data for the month %d: ", j + 1);
      scanf("%f", &arr_brms[i][j]);
    }
    printf("\n");
  }
}

float sum_of_sales(int number_of_branches ,float arr_brms[number_of_branches][12],float total_sale)
{
  total_sale = 0;
  for (int i = 0; i < number_of_branches; i++)
    for (int j = 0; j < 12; j++) total_sale += arr_brms[i][j];
  printf("The sum of all the branches income combined is = %f\n\n", total_sale);
return total_sale;}

void percentage(int number_of_branches,float arr_brms[number_of_branches][12],float total_sale)
{
int i ,j, m;
    float total = 0;
    total = total_sale;
   for(i=0;i<number_of_branches;i++){
      for(j=0;j<m;j++){
        total = total + arr_brms[i][j];}
                printf("Branch %d : ",i+1);
    float per , sum ;
    for(i=0;i<number_of_branches;i++){
      sum=0;
      per=0;
    }
      for(j=0;j<m;j++){
      sum+= arr_brms[i][j];

        per = (sum / total) * 100.0;
        printf("Percentage = %.2f%%",per);
        printf("\n") ;
    }
}
    }

int main()
{
  printf(
      "Please enter the number of branches you'd like to input data about: ");
  int number_of_branches = 0;
  scanf("%d", &number_of_branches);
  float arr_brms[number_of_branches][12];
  Data_Of_Branches(number_of_branches, arr_brms);
  printf("Hello. please enter the number of the operation you'd like to do. "
         "(please only enter the number.) \n \n");
  int choice;
    float total_sale;
  while (
    printf(" 1. Enter sales data.\n 2. Add a record for a new branch \n 3. "
           "Delete record of an existing branch \n 4. Calculate total sales \n "
           "5. Calculate percentage share of each branch \n 6. Determine the "
           "month of the peak sales \n 7. Display sales of a specific month \n "
           "8. Display sales of a specific branch \n 0. Done \n"),
    scanf("%d", &choice) > 0
        )
  {
    if (choice == 0) exit(0);
    if (choice == 1) Data_Of_Branches(number_of_branches, arr_brms);
    if (choice == 4) sum_of_sales(number_of_branches, arr_brms,total_sale);
        if (choice == 5)percentage(number_of_branches, arr_brms,total_sale);
  }
  return 0;
}

编辑：好的，所以基本上我需要的是解决sum函数的方法。因为它一直显示错误的数字。另一个函数显示总和中每个分支的百分比编辑2：好的，所以在用Armali放的新代码替换旧代码之后。我继续前进，开始失败-添加百分比函数失败。因此，如果可能的话，有一种方法可以修复空隙百分比函数？