时间限制 1000 ms 内存限制 65536 KB 代码长度限制 100 KB 判断程序 Standard (来自 小小) 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 M7) to get M28 back; coupon 2 to product 2 to get M12 back; and coupon 4 to product 4 to get M3 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
输入描述: 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.
输出描述: For each test case, simply print in a line the maximum amount of money you can get back.
输入例子: 4 1 2 4 -1 4 7 6 -2 -3
输出例子: 43
题解：给出两个集合，然后选取元素使乘积最大，可以想到贪心策略，把两个数组从小到大排序， 然后从左到右选取相同位置都是负数相乘，从右到左选取相同位置非负数相乘，然后便可以得到最大值。 平台是牛客，数据比较大，记得使用long long 类型。
#include <bits/stdc++.h> #include <stdlib.h> using namespace std; #define vi vector<int> #define pii pair<int,int> #define x first #define y second #define all(x) x.begin(),x.end() #define pb push_back #define mp make_pair #define SZ(x) x.size() #define rep(i,a,b) for(int i=a;i<b;i++) #define per(i,a,b) for(int i=b-1;i>=a;i--) #define pi acos(-1) #define mod 1000000007 #define inf 1000000007 #define ll long long #define DBG(x) cerr<<(#x)<<"="<<x<<"\n"; #define N 200010 template <class U,class T> void Max(U &x, T y){if(x<y)x=y;} template <class U,class T> void Min(U &x, T y){if(x>y)x=y;} template <class T> void add(int &a,T b){a=(a+b)%mod;} const int maxn=100001; int n,m; ll a1[maxn],a2[maxn]; int main() { cin>>n; rep(i,0,n) scanf("%lld",&a1[i]); cin>>m; rep(i,0,m) scanf("%lld",&a2[i]); sort(a1,a1+n); sort(a2,a2+m); int i=0,j=0; ll ans=0; while(a1[i]<0&&a2[i]<0&&i<n&&i<m) { ans+=a1[i]*a2[i]; i++; } i=n-1,j=m-1; while(a1[i]>0&&a2[j]>0&&i>=0&&j>=0) { ans+=a1[i]*a2[j]; i--,j--; } printf("%lld",ans); return 0; }