The magic shop in Mars is offering some magic coupons. Each coupon has an integer Nprinted 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 − }, and a set of product values { 7 6 − − } (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, 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
这题只能说是简单贪心了
1 #include <bits/stdc++.h> 2 #define N 100005 3 using namespace std; 4 int n,m; 5 int an[N], bn[N]; 6 7 int main(){ 8 cin >> n; 9 for(int i = 0; i < n; i++){ 10 cin >> an[i]; 11 } 12 sort(an,an+n); 13 cin >> m; 14 for(int i = 0; i < m; i++){ 15 cin >> bn[i]; 16 } 17 sort(bn,bn+m); 18 int sum = 0; 19 int i = 0, ii = n-1; 20 int j = 0, jj = m-1; 21 while(i <= ii && j <= jj){ 22 bool flag = true; 23 if(an[i]*bn[j] > an[ii]*bn[jj]){ 24 if(an[i]*bn[j]>0){ 25 sum += an[i]*bn[j]; 26 i++; 27 j++; 28 flag = false; 29 } 30 }else{ 31 if(an[ii]*bn[jj] > 0){ 32 sum += an[ii]*bn[jj]; 33 ii--; 34 jj--; 35 flag = false; 36 } 37 } 38 if(flag) 39 break; 40 } 41 cout << sum << endl; 42 return 0; 43 }