【C++】PAT(advanced level)1037. Magic Coupon (25)

1037. Magic Coupon (25)

时间限制
100 ms

内存限制
32000 kB

代码长度限制
16000 B

判题程序
Standard

作者
CHEN, Yue

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:

4
1 2 4 -1
4
7 6 -2 -3

Sample Output:

43

一次AC，主要是读懂题意

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<map>
#include<iomanip>
using namespace std;

bool cmp(int a,int b){
return a>b;
}

int main(){
//freopen("in.txt","r",stdin);
int NC,NP;
cin>>NC;
int n=NC,m;
vector<int> nc;
vector<int> np;
while(n--){
int temp;
scanf("%d",&temp);
nc.push_back(temp);
}
sort(nc.begin(),nc.end(),cmp);
cin>>NP;
m=NP;
while(m--){
int tt;
scanf("%d",&tt);
np.push_back(tt);
}
sort(np.begin(),np.end(),cmp);
n=np.size();
int i=0,sum=0;
while(true){
if(nc[i]>0&&np[i]>0){
sum+=nc[i]*np[i];
i++;
}else{
break;
}
}
i=1;
while(true){
if(nc[NC-i]<0&&np[NP-i]<0){
sum+=nc[NC-i]*np[NP-i];
i++;
}else{
break;
}
}
cout<<sum;
//fclose(stdin);
system("pause");
return 0;
}