纸币平均数 C++实现

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#C++ #程序设计 #穷举法 #贪心法

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本文探讨了两种不同的算法——穷举法和贪心法，在解决特定纸币找零问题中的应用。通过具体实例展示了每种算法如何计算购买不同价格商品所需的最少纸币数量，并比较了它们在实际操作中的平均效率。

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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 <iostream>
#include <ctime>
#include <cstdlib>
#include <cmath>

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 <iostream>
#include <ctime>
#include <cstdlib>
#include <cmath>

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;
}