PAT A1070 月饼题

PAT A1070 月饼题

这道题个人觉得不严谨……
如果所有月饼的库存量都满足不了需求量该怎么输出没有提……（如果我眼花了请指正）
测试点3我的出现了段错误，找半天发现是while循环中i的限制，原本是这样写的：

```cpp
while (true)
	{
		gram = gram + cookie[i].kucun;
		money = money + cookie[i].all;
		if (gram >= d) {
			money = money - (gram - d) * cookie[i].price;
			break;
		}
		i++;
	}

后来改成：

```cpp
while (true)
	{
		gram = gram + cookie[i].kucun;
		money = money + cookie[i].all;
		if (gram >= d) {
			money = money - (gram - d) * cookie[i].price;
			break;
		}
		i++;
        if(i == n) break;////////加了这句
	}

限制了i以后测试点3的段错误就过了……
下面是完整的代码：

#include<iostream>
#include <string>
#include <cstdio>
#include<cmath>
#include<algorithm>//min()、sort()
#include<iomanip>
#include<map>
using namespace std;
struct moon {
	double kucun;
	double all;
	double price;
};
moon cookie[1010];
bool cmp(moon a, moon b) {
	return a.price > b.price;
 }

int main() {
	int n;
	double d;
	cin >> n;
	cin >> d;
	for (int i = 0; i < n; i++)
	{
		cin >> cookie[i].kucun;
	}
	for (int i = 0; i < n; i++)
	{
		cin >> cookie[i].all;
		cookie[i].price = cookie[i].all / cookie[i].kucun;
	}

	sort(cookie, cookie + n, cmp);
	double gram = 0;
	double money = 0;
	int i = 0;
	while (true)
	{
		gram = gram + cookie[i].kucun;
		money = money + cookie[i].all;
		if (gram >= d) {
			money = money - (gram - d) * cookie[i].price;
			break;
		}
		i++;
        if(i == n) break;
	}

	cout << fixed<<setprecision(2) <<  money;

	
}