钱币找零问题-贪心算法解析

#include <iostream> 

using namespace std;

#define N  7 

int counts[N] = { 3, 2, 0, 3 ,2 ,0 , 1 };
int value[N] = { 1, 2, 5, 10, 20, 50, 100 }; 
int cot[N] = { 0, 0, 0, 0 ,0 ,0 , 0 };

int solve(int money) {
	int num = 0; //支付了几张纸币
	int i = 0;

	for (i = N - 1; i >= 0; i--) {
		int j = money / value[i]; //对应的纸币要给几张
		//有这么多给对应的钱,没有那些有多少给多少
		int c = j > counts[i] ? counts[i] : j; 

		cout << "需要面值" << value[i] << "的纸币" << c << "张" << endl;
		money -= c * value[i]; //还剩多少要给
		num += c; 

		if (money == 0) { break; }
	}
	if (money > 0) { num = -1; } //不能完全支付

	return num;
}

int main() {
	int money = 0;
	int num = 0;

	cout << "请输入要支付的数目:\n";
	cin >> money;

	num = solve(money);

	if (num == -1) {
		cout << "支付失败!" << endl;
	}
	else {
		cout << "成功支付款项:" << num << endl;
	}

	system("pause");
	return 0;
}