C语言实现：将 8 - 100 零钱换 5、2、1 分硬币的所有换法揭秘

```c #include <stdio.h> void change(int money) { int count = 0; for (int i = money / 5; i >= 1; i--) { for (int j = (money - 5 * i) / 2; j >= 1; j--) { for (int k = money - 5 * i - 2 * j; k >= 1; k--) { if (5 * i + 2 * j + k == money) { printf("fen5:%d, fen2:%d, fen1:%d, total:%d\n", i, j, k, i + j + k); count++; } } } } printf("count = %d\n", count); } int main() { int money; scanf("%d", &money); change(money); return 0; } ``` ######[AI写代码神器 | 216点数解答 | 2024-10-31 09:33:55]