PAT B1020

PAT B1020

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解决思路 :贪心法，每次选取单价最高的月饼。

先上一个自己错误的解法

#include <cstdio>
#include <algorithm>
using namespace std;

double num[1010];
double price[1005];
double ans = 0;

int main() {
    int n, d;
    scanf("%d%d", &n, &d);
    for (int i = 0; i < n; i++) {
        scanf("%lf", &num[i]);
    }
    for (int i = 0; i < n; i++) {
        scanf("%lf", &price[i]);
        price[i] = price[i] / num[i];
    }
    sort(price, price + n);

    for (int i = n - 1; i >= 0; i--) {
        while (num[i] > 0 && d > 0) {
            ans += price[i];
            num[i]--;
            d--;
        }
    }


    printf("%.2f", ans);
    return 0;
}

然后是题解

#include <cstdio>
#include <algorithm>
using namespace std;

struct mooncake {
    double store; //库存
    double sell;   //总价
    double price;  /单价
}cake[1010];

bool cmp(mooncake a, mooncake b) {
    return a.price > b.price;
}

int main() {
   int n;
   double D;
   scanf("%d%lf", &n, &D);
   for (int i = 0; i < n; i++) {
    scanf("%lf", &cake[i].store);
   }
   for (int i = 0; i < n; i++) {
    scanf("%lf", &cake[i].sell);
    cake[i].price = cake[i].sell / cake[i].store;
   }
   sort(cake, cake + n, cmp);
   double ans = 0;
   for (int i = 0; i < n; i++) {   
    if (cake[i].store <= D) {    //如果需求高于月饼的库存
        D -= cake[i].store;    //先把第i种全部卖出
        ans += cake[i].sell;
    } else {                 //库存高于需求
        ans += cake[i].price * D;   //只售出该种月饼，数量为需求量，然后break；
        break;
    }
   }
   printf("%.2f", ans);
   return 0;
}