纸币平均数 C++实现

购买某种价格的物品最少需要几张人民币？

纸币一共有五种：
1元、5元、10元、20元和50元

要购买物品的价格为1～99元的整数：

① 购买每种价格的物品最少需要几张纸币？

② 随机购买一种物品，平均最少要几张纸币？

下面分别用 穷举法 和 贪心法 求解 ：

（1）穷举法：

#include 
#include 
#include 
#include 

using namespace std;

int main() {
    //i:循环变量; n:循环次数; money:当前的物品价格;
    //firstone~fifthone:对应纸币的张数;
    int i, n, money, o, sum, average=0, firstone, secondone, thirdone, fourthone, fifthone, p, q, r, s, t, minimum;
    //设置时间为随机数的种子
    srand(time(NULL));

    cout << "input five types of money: " << endl;
    cout << "(increasing order please)" << endl;
    cin >> p >> q >> r >> s >> t;

    cout << "input test times:" << endl;
    cin >> n;

    for (i = 1; i <= n; i++) {
        //用随机数映射到1~100作为当前商品的价格
        money = rand() * 100 / (RAND_MAX +1) + 1;
        o = money;
        //设置初始值
        sum = 0;
        minimum = 1000;
        firstone = 0;
        secondone = 0;
        thirdone = 0;
        fourthone = 0;
        fifthone = 0;
        //用for循环嵌套的方式把所有的符合总价的纸币组合列出来，计算张数，每次计算出来，判断是不是最小张数，然后更新
        for (firstone = 0; firstone <= (money / t); firstone++) {
            for (secondone = 0; secondone <= (o - firstone * t); secondone++) {
                for (thirdone = 0; thirdone <= (o - firstone * t - secondone * s); thirdone++) {
                    for (fourthone = 0; fourthone <= (o - firstone * t - secondone * s - thirdone * r); fourthone++) {
                        for (fifthone = 0; fifthone <= (o - firstone * t - secondone * s - thirdone * r -
                                                        fourthone * q); fifthone++) {
                            if (firstone * t + secondone * s + thirdone * r + fourthone * q + fifthone * p == money) {
                                sum = firstone + secondone + thirdone + fourthone + fifthone;
                                if (minimum > sum) minimum = sum;
                                //注意每算完一种价值的物品，相关变量都要清零！
                                sum = 0;
                                o = money;
                            }
                        }
                    }
                }
            }
        }
        average += minimum;
        cout << money << "    " << minimum << endl;
    }
    cout << (average / n) << endl;
    return 0;
}

（2）贪心法：

#include 
#include 
#include 
#include 

using namespace std;

int main() {
	//	变量名含义与穷举法相同
    int i, n, money, o, sum, average, firstone, secondone, thirdone, fourthone, fifthone, p, q, r, s, t;

    average = 0;
    srand(time(NULL));

    cout << "input five types of money: " << endl;
    cout << "(increasing order please)" << endl;
    cin >> p >> q >> r >> s >> t;

    cout << "input test times:" << endl;
    cin >> n;

    for (i = 1; i <= n; i++){
        money = rand()*100/(RAND_MAX+1)+1;
        o = money;
        sum = 0;
        //用来记录每种纸币的张数
        firstone = 0;
        secondone = 0;
        thirdone = 0;
        fourthone = 0;
        fifthone = 0;
        //面值从大到小考虑
        if (o >= t) {
            fifthone = o / t;
            o %= t;
        }
        if (o >= s) {
            fourthone = o / s;
            o %= s;
        }
        if (o >= r) {
            thirdone = o / r;
            o %= r;
        }
        if (o >= q) {
            secondone = o / q;
            o %= q;
        }
        if (o >= p) {
            secondone = o / p;
            o %= p;
        }

        sum = firstone + secondone + thirdone + fourthone + fifthone;
        average += sum;

        cout << money << " yuan's products need " << sum << " pieces of RMB" << endl;
    }
    cout << (average / n) << endl;

    return 0;
}

运行结果

① 如价格为 67 元，则需要 5 张纸币；
② 随机购买 1000 次，平均需要 4 张纸币；

思考：

① 两种方法的结果是否总是相同？
（试输入纸币的种类为1元、5元、16元、23元和33元）

② 贪婪法和穷举法的各自优劣？