找零問題-徹底揹包

問題描述

100元錢，有幾種零錢，好比1元，5元，10元，求有幾種組合

分析

時間複雜度O(m*n)，空間複雜度O(n)，徹底揹包問題

代碼以下

 1 #include <stdio.h>
 2 #define N 100
 3 int a[N + 5]; 
 4 int change[4]; //有幾種零錢
 5 //主功能函數，找到一共須要幾種零錢
 6 int findCount(int money, int countOfChange)
 7 {
 8     int i, j;
 9     a[0] = 1;
10     for (i = 1; i <= countOfChange; i++)
11     {
12         for (j = change[i]; j <= money; j++)
13         {
14             //核心的代碼，就此一句
15             a[j] = a[j] + a[j - change[i]];
16         }
17         /*
18         //去掉註釋，可打印每一步的狀況
19         for (j = 0; j <= money; j++)
20             printf("%d\t", a[j]);
21         puts("");*/
22         
23     }
24     return a[money];
25 }
26 int main()
27 {
28     int money, num, count, i;
29     money = 6;
30     num = 3;
31     //一系列輸入
32     printf("input the money u have: ");
33     scanf("%d", &money);
34     printf("input the kinds of changes: ");
35     scanf("%d", &num);
36     for (i = 1; i <= num; i++)
37     {
38         printf("the %dth change: ", i);
39         scanf("%d", &change[i]);
40     }
41     count = findCount(money, num);
42     printf("the count is %d.\n", count);
43     return 0;
44 }