# PAT 解题报告 1070. Mooncake (25)

### 1070. Mooncake题目描述：

Mooncake is a Chinese bakery product traditionally eaten during the Mid-Autumn Festival. Many types of fillings and crusts can be found in traditional mooncakes according to the region’s culture. Now given the inventory amounts and the prices of all kinds of the mooncakes, together with the maximum total demand of the market, you are supposed to tell the maximum profit that can be made.

Note: partial inventory storage can be taken. The sample shows the following situation: given three kinds of mooncakes with inventory amounts being 180, 150, and 100 thousand tons, and the prices being 7.5, 7.2, and 4.5 billion yuans. If the market demand can be at most 200 thousand tons, the best we can do is to sell 150 thousand tons of the second kind of mooncake, and 50 thousand tons of the third kind. Hence the total profit is 7.2 + 4.5/2 = 9.45 (billion yuans).

### 1070. Mooncake算法分析：

```#include <cstdio>
#include <algorithm>
#include <vector>

struct Node {
double amount;
double price;
};

bool cmp(const Node &n1, const Node &n2) {
return n1.price * n2.amount > n2.price * n1.amount;
}

double gao(int n, double d, std::vector<Node> &cakes) {
std::sort(cakes.begin(), cakes.end(), cmp);
double profit = 0.0;
for(int i = 0; d > 0 && i < n; ++i) {
int sell = std::min(d, cakes[i].amount);
profit += cakes[i].price * (sell * 1.0 / cakes[i].amount);
d -= sell;
}
return profit;
}

int main() {
int N, D;
scanf("%d %d", &N, &D);

std::vector<Node> cakes(N);
for(int i = 0; i < N; ++i)
scanf("%lf", &cakes[i].amount);
for(int i = 0; i < N; ++i)
scanf("%lf", &cakes[i].price);

printf( "%.2lf\n", gao(N, D, cakes) );
return 0;
}
```

### 1070. Mooncake注意点：

Written on November 30, 2013