PAT 解题报告 1037. Magic Coupon (25)

1037. Magic Coupon 题目描述:

The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the shop also offers some bonus product for free. However, if you apply a coupon with a positive N to this bonus product, you will have to pay the shop N times the value of the bonus product… but hey, magically, they have some coupons with negative N’s!

For example, given a set of coupons {1 2 4 -1}, and a set of product values {7 6 -2 -3} (in Mars dollars M$) where a negative value corresponds to a bonus product. You can apply coupon 3 (with N being 4) to product 1 (with value M$7) to get M$28 back; coupon 2 to product 2 to get M$12 back; and coupon 4 to product 4 to get M$3 back. On the other hand, if you apply coupon 3 to product 4, you will have to pay M$12 to the shop.

Each coupon and each product may be selected at most once. Your task is to get as much money back as possible.

题目背景是给你一堆Coupon和一堆Product, 给你设定了一些规则, 求如何分配这些Coupon使得收益率最大, 回收的money最多. 抽象出来就是一个数学题目: 给你两个数组A和B, 满足max_{i, j} \sigma {A[i] * B[j]}, 也就是, 分配一个coupon A[i]给一个product B[j]我们可以得到的money是A[i] * B[j],  如何分配使得这些个A[i] *B[j] 的总和最大.

1037. Magic Coupon 算法分析:

算法属于贪心算法吧. 给Coupon和Product都统一从大到小排序(或者从小到大排序), 然后检测两个数组的头尾, 头尾都扫一遍, 把所有乘积是正的A[i] * B[j] 都加到最终的结果中就可以了. 代码如下:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <iomanip>

#define nmax 100005
using namespace std;

int NC;
int NP;
int minimum;
long long money;

int coupons[nmax];
int products[nmax];

int compare(const void * a, const void * b) { return -( *(int*)a - *(int*)b ); }

void gao(void) {
    int count = 0;
    int c = 0;
    long long prod = coupons1*products1;
    while(prod > 0 && count<minimum) {
        money += prod;
        prod = coupons1*products1;
    prod = coupons[NC-c]*products[NP-c];
    while(prod>0 && count<minimum) {
        money += prod;
        prod = coupons[NC-c]*products[NP-c];
    return ;

int main(void) {
    for(int i=0; i<NC; i++) cin>>coupons[i];
    for(int i=0; i<NP; i++) cin>>products[i];
    money = 0;
    minimum = min(NC,NP);
    return 0;

1037. Magic Coupon 注意点:

扫的过程中注意不要重复加入同一个A[i] * B[j] 代码中稍微做点判断应该没什么问题. 参考上面的源代码.

Written on December 29, 2013