硬币找零问题中要求找给客户最少的硬币。coins存储可用硬币规格，单位为角，假设规格都小于10角，且一定有1角规格。amount为要找零的金额，约定必须为1角的整数倍。输出为每种规格及其数量，按规格从大到小输出，如果某种规格不必要，则输出为0。下面是其实现代码：

const int MAX_COINS = 10;
int result[MAX_COINS] = {0}; // 假设最多10种面额
int find_coins(const vector<int>& coins, int amount) {
    sort(coins.begin(), coins.end(), greater<int>());
    int n = coins.size();
    for (int i = 0; i < n; ++i) {
        int coin = coins[i];
        int num = amount / coin;
        result[i] = num;
        amount -= num * coin;
        if (amount == 0) break;
    }
    cout << "找零方案如下：" << endl;
    for (int i = 0; i < n; ++i) {
        cout << sorted_coins[i] << "角需要" << result[i] << "枚" << endl;
    }
    return 0;
}