Question

#include <stdio.h>

void main()
{
    int count = 0, amount, temp, n_200, n_100, n_50, n_20, n_10, n_5, n_2, n_1;    /*count=check how much posibiliti you have,amount=the amount that the user enter   ,n_200=bill of 200 nis ,n_100=bill of 100 nis.... n_1=1 nis coin.*/
    double min = 500, max = 0, current = 0;    
//part 2 of the job check the max and min whight...

    while (1) {                         //the progrem run all the time
        printf_s("What is the amount that you like to check? (or press '0' to exit)\n");
        scanf_s("%d", &amount);
        temp = amount;
        if (amount == 0)
//if the user click 0 the lop is finish and the progrem exit
            return;

        for (n_200 = 0; amount - n_200 >= 0; n_200 += 200)    
//first lop the duplicate of 200
        {

            for (n_100 = 0; amount - n_100 >= 0; n_100 += 100)   
//secend lop the duplicate of 100
            {


                for (n_50 = 0; amount - n_50 >= 0; n_50 += 50)     
//thired lop the duplicate of 50
                {


                    for (n_20 = 0; (amount - n_20) / 20 >= 0; n_20 += 20)     //fourth lop the duplicate of 20
                    {


                        for (n_10 = 0; amount - n_10 >= 0; n_10 += 10)   
//fifth lop the duplicate of 10
                        {


                            for (n_5 = 0; amount - n_5 >= 0; n_5 += 5)     
//six lop the duplicate of 5  
                            {


                                for (n_2 = 0; amount - n_2 >= 0; n_2 += 2)    //seventh lop the duplicate of 2  
                                {


                                    for (n_1 = 0; amount - n_1 >= 0; n_1 += 1)  //eighth lop the duplicate of 1  
                                    {
                                        temp = amount - n_200 - n_100 - n_50 - n_20 - n_10 - n_5 - n_2 - n_1;


                                        if (temp == 0) {  /////if temp==0 all the money is returend,the counter (posibiliti) is grow by one and the loop start again this time with other posibiliti.
                                            count += 1;

                                            current = current + (n_200 / 200 + n_100 / 100 + n_50 / 50 + n_20 / 20)*0.001 + (n_10 / 10)*0.007 + (n_5 / 5)*0.0082 + (n_2 / 2)*0.0057 + n_1*0.004;  
// In each variation we check the whight (curent) 
                                            if (current > max)   
//if the current is bigerr the max 
                                                max = current;   
//the progrem save it as max
                                            if (current < min)   
//if the current is smaller the min
                                                min = current;   
//the progrem save it as min`enter code here`


                                            temp = amount;
                                            current = 0;
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
        printf_s("\nThe number of possibilities is: %d.\n", count);
        printf_s("The maximum weight is %.4lf [kg]\n", max);
        printf_s("The minimmum weight is %.4lf [kg]\n\n", min);
        count = 0;                 //to start again we rest the deta the we get.
        min = 500, max = 0, current = 0;
    }
}

我需要计算1,2,5,10,20,50,100,200之和的组合数量，以便用户加起来给定数量。例如，如果用户输入5，则组合可以是：

1+1+1+1+1
1+1+1+2
2+2+1
5

Answer 1

蛮力：

#include <stdio.h>

/* coin_types must be sorted from smallest to largest */
int count_coin_combinations(const unsigned *coin_types, unsigned num_coin_types, unsigned n) {
   if (n == 0)
      return 1;

   int count = 0;
   while (num_coin_types && *coin_types <= n) {
      count += count_coin_combinations(coin_types, num_coin_types, n-*coin_types);
      ++coin_types;
      --num_coin_types;
   }

   return count;
}

int main() {
   const unsigned coin_types[] = { 1, 2, 5, 10, 20, 50, 100, 200 };
   const unsigned num_coin_types = sizeof(coin_types)/sizeof(*coin_types);

   unsigned n = 5;
   unsigned count = count_coin_combinations(coin_types, num_coin_types, n);
   printf("%d\n", count);

   return 0;
}

输出：

$ gcc -Wall -Wextra -pedantic -std=c99 -o a a.c && a
4

如何更有效地计算组合数？

1 个答案: