题目：给定一个金额m，以及几种钱币面值（比如1，2，5），求m有多少种找钱方式

解答：

a(m, c): 金额m的找钱方式，此时最大钱币面值为c

a(m, c) = sigma( a(m - k*c, next_c) ); k=0~m/c, next_c=比c小的下一个面值的钱币，比如c=5, next_c = 2

按照以上递推式，可以写出递归函数：

int exchangeWays(int money, int maxCurIndex = currencyNum - 1)
{
    if (money == 0) return 1;
    if (maxCurIndex == -1) return 0;
    int ways = 0;
    int maxCur = currency[maxCurIndex];
    for (int i = 0; i * maxCur <= money; i++)
        ways += exchangeWays(money - i * maxCur, maxCurIndex - 1);
    return ways;
}

循环解法1：

有几个c就有几层循环，复杂度也是O(m^(c的个数))

int exchangeWaysIter(int money)
{
    int ways = 0;
    for (int i = 0; i * currency[2] <= money; i++)
    {
        money -= i * currency[2];
        for (int j = 0; j * currency[1] <= money ; j++)
        {
            money -= j * currency[1];
            for (int k = 0; k * currency[0] <= money; k++)//if currency[0] is 1, we can delete this loop
            {
                if (money == k * currency[0])
                    ++ways;
            }
            money += j * currency[1];
        }
        money += i * currency[2];
    }
    return ways;
}

int exchangeWaysIter2(int money)
{
    vector<int> ways(money + 1);//ways(i) : sum equals i, exchange ways
    ways[0] = 1;
    for (int i = 0; i < currencyNum; i++)
    {
        for (int j = 0; j + currency[i] <= money; j++)
        {
            if (ways[j])
                ways[j + currency[i]] += ways[j];
        }
    }
    return ways[money];
}

一下就降低了时间复杂度。

找钱方式:递归，循环的解法