PAT刷题：1037. Magic Coupon (25)

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.

Input Specification:

Each input file contains one test case. For each case, the first line contains the number of coupons NC, followed by a line with NC coupon integers. Then the next line contains the number of products NP, followed by a line with NP product values. Here 1<= NC, NP <= 10⁵, and it is guaranteed that all the numbers will not exceed 2³⁰.

Output Specification:

For each test case, simply print in a line the maximum amount of money you can get back.

Sample Input:

41 2 4 -147 6 -2 -3

Sample Output:

43

#include <iostream>#include <string>#include <vector>#include <algorithm>#include <cstdio>using namespace std;bool cmp(const int &a , const int &b){return a>b;}int main(){    int m=0 , n=0;    vector<int>coupons , products;    cin>>m;    coupons.resize(m);    for(int i=0;i<m;i++)cin>>coupons[i];    cin>>n;    products.resize(n);    for(int i=0;i<n;i++)cin>>products[i];    sort(coupons.begin() , coupons.end() , cmp);    sort(products.begin() , products.end() , cmp);    int total=0;    while(coupons.size()>0 && products.size()>0){        if(coupons.front() > 0 && products.front() >0){            total+=coupons.front() * products.front();            coupons.erase(coupons.begin());            products.erase(products.begin());        }        else if(coupons.back() < 0 && products.back() <0){            total+=coupons.back()*products.back();            coupons.pop_back();            products.pop_back();        }        else break;    }    cout<<total;    return 0;}

#include <iostream>#include <string>#include <vector>#include <algorithm>#include <cstdio>using namespace std;bool cmp1(const int &a , const int &b){return a>b;}bool cmp2(const int &a , const int &b){return a<b;}int main(){    int m=0 , n=0;    vector<int>p_coupons , p_products ,n_coupons , n_products;    cin>>m;    p_coupons.resize(m);    n_coupons.resize(m);    for(int i=0;i<m;i++){        int t=0;        cin>>t;        if(t>0)p_coupons.push_back(t);        else n_coupons.push_back(t);    }    cin>>n;    p_products.resize(n);    n_products.resize(n);    for(int i=0;i<n;i++){        int t=0;        cin>>t;        if(t>0)p_products.push_back(t);        else n_products.push_back(t);    }    sort(p_coupons.begin() , p_coupons.end() , cmp1);    sort(p_products.begin() , p_products.end() , cmp1);    sort(n_coupons.begin() , n_coupons.end() , cmp2);    sort(n_products.begin() , n_products.end() , cmp2);    int total=0;    int _min=std::min(p_coupons.size() , p_products.size());    for(int i=0;i<_min;i++){        total+=p_coupons[i]*p_products[i];    }    _min=std::min(n_coupons.size() , n_products.size());    for(int i=0;i<_min;i++){        total+=n_coupons[i]*n_products[i];    }    cout<<total;    return 0;}